Higher Mathematics for Students of Chemistry and Physics

Higher Mathematics for Students of Chemistry and Physics Joseph William Mellor 1869-1938 Chapter VII. How to Solve Differential Equations Preface to 2009 electronic edition THE PRIMARY content of this e-book is chapter VII of the 1902 edition of Higher Mathematics for Students of Chemistry and Physics (with special reference to practical work) by Joseph William Mellor, D.Sc of New Zealand Univerity (published by Longmans, Green, and Company of London). Chapter VII of this text is an 80 page missive entitled, “How to Solve Differential Equations.” The text is primarily concerned with teaching practical methods for arriving at solutions, and contains almost nothing on the theory of differential equations – that is, there are very few theorems and proofs. It covers a wide variety of solvable ordinary differential equations with over a hundred examples with solutions. The later sections also provide an introductory overview of partial differential equations. Included also is a supplement I have prepared with many of the examples in the main text worked out step-by-step. Some of them I have worked in more than one way. Students using this material are encouraged to attempt each example in the main text on their own before referring to the supplement for my working of that example. In addition the supplement contains my comments on the main text, as well as descriptions of methods of solution not covered in the main text. No student should expect to learn the material here well without working the examples for himself or herself. Prerequisite skills: Students using this material should be competent in finding derivatives, including applying the product, quotient, and chain rules, in methods for finding integrals, and in the manipulation of polynomials, roots, logs, exponentials, and trigonometric functions. Students should also have had at least some exposure to the differential calculus of functions of more than one variable and partial derivatives. The text makes scattered use of complex numbers, as does the supplement. So students should be familiar with the algebra of complex numbers. They should also be familiar with Euler’s formula relating exponentials of a complex variable to the sine and cosine functions: eιx + e−ιx eιx − e−ιx eιx = cos x + ι sin x hence cos x = and sin x = . 2 2ι i

√ Note the usage of the symbol, ι, for −1. This notation is used throughout the text. Most modern texts use the symbol, i, in lieu of ι for this purpose. I have continued Mellor’s notation, using ι in the supplement in order to remain consistent. Mellor uses the adjective, “imaginary,” to describe any non-real complex number. This is at odds with modern nomenclature, in which “imaginary” refers only to real multiples of ι. In the supplement, I use the phrase, “purely imaginary,” to indicate such multiples and to make the distinction from Mellor’s looser usage. More on notation: Throughout the text and the supplement, the notation, log (x), is always to be understood as logarithm to the base, e, otherwise known as the natural log. Most modern texts use the notation, ln (x), for natural log. The website, http://jeff560.tripod.com/functions.html, indicates that the ln (x) notation was not introduced until 1893, so at the time of publication of this text, that notation was still new and not yet widely adopted. The log (x) notation, according to the same source, had been in use since 1748. At any rate, the text uses the older notation, and I have stuck with it in the supplement. The text consistently uses the following hierarchy for grouping of terms: [{()}]. Again I have stuck with that style in the supplement. The original text used an unfamiliar notation for limits. Since its notation is no longer in common use, I have converted it to the modern notation, e.g., sin (h) lim = 1. h→0 h Errors in the original text: In preparing the electronic version of the text from the original printed material (and especially when I prepared the supplement), I discovered many errors in the text. Some of these were type-setting errors, but others were the result of carelessness on the part of the author. Where mistakes were obvious, I have simply corrected them without comment. Where there were mistakes of carelessness in the examples and their solutions, I have, in many cases, left the mistake unaltered, but I have tagged it in the right-hand margin with a left-facing arrow. Where you encounter the arrow, you will find my reasons for believing that the text is in error in the supplement, together with what I believe the text should have been. Serious students will attempt to find the error on their own before referring to the supplement. In many other places I have also inserted comments in the right-hand margin, many of which refer the student to the supplement for further clarification. The page numbering of the original printed text is preserved in this electronic version. The supplement uses its own page numbering, starting with S-1, which follows page 359 of the main text. Following that are various excerpts from other chapters of the main text that are referred to in chapter VII. The supplement includes three final sections that are not specifically referenced from example problems in the main text. These are Resonant systems revisited starting on page S-61, which describes how resonant ii

systems respond to sinusoidal input, Differential equation solution to Keplerian orbits starting on page S-66, which develops the shape of a gravitational orbit in a two body problem, and Introduction to Fourier’s heat equation and its solution starting on page S-67, which develops methods for dealing with a one-dimensional heat equation. These have been added because they are important applications of the methods taught in this tutorial and may be of interest to some students. The text by J.W. Mellor passes into public domain on the first day of the year 2009. My comments and supplement are offered without charge to all under Creative Commons 3.0 license. See http://creativecommons.org/licenses/by/3.0/ for details. I am often receiving emails from students living in Africa, the Middle East, or South Central Asia, who are eager to learn mathematics but for whom the cost of modern textbooks is burdensome. One of the reasons I have gone to the effort to prepare an electronic edition of this material is for them. If you are a math educator working in those parts of the world and teaching differential equations, please consider using this material as part of your curriculum so as to allow your students affordable access to the knowledge it provides. LATEX sources for this material will be made available, free of charge, to anybody who can demonstrate that he or she will put them to good use. Translation of the material into languages other than English will always be considered to be a good use. Email me for details. The probability that I have transcribed all of this material without error is vanishingly close to zero. If you discover anything in this document that appears to you to be in error, please notify me by email so that I may correct it. Karl Hahn, December 2008

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Mellor’s Preface to the 1902 edition IT is almost impossible to follow the later developments of physical or general chemistry without a working knowledge of higher mathematics. I have found that the regular textbooks of mathematics rather perplex than assist the chemical student who seeks a short road to this knowledge, for it is not easy to discover the relation which the pure abstractions of formal mathematics bear to the problems which every day confront the student of Nature’s laws, and realize the complementary character of mathematical and physical processes. During the last five years I have taken note of the chief difficulties met with in the application of the mathematicians x and y to physical chemistry, and, as these notes have grown, I have sought to make clear how experimental results lend themselves to mathematical treatment. I have found by trial that it is possible to interest chemical students and to give them a working knowledge of mathematics by manipulating the results of physical or chemical observations. I should have hesitated to proceed beyond this experimental stage if I had not found at The Owens College a set of students eagerly pursuing work in different branches of physical chemistry, and most of them looking for help in the discussion of their results. When I told my plan to the Professor of Chemistry he encouraged me to write this book. It has been my aim to carry out his suggestion, so I quote his letter as giving the spirit of the book, which I only wish I could have carried out to the letter. “THE OWENS COLLEGE, MANCHESTER.”

“MY DEAR MELLOR, “If you will convert your ideas into words and write a book explaining the inwardness of mathematical operations as applied to chemical results, I believe you will confer a benefit on many students of chemistry. We chemists, as a tribe, fight shy of any symbols but our own. I know very well you have the power of winning new results in chemistry and discussing them mathematically. Can you lead us up the high hill by gentle slopes? Talk to us chemically to beguile the way? Dose us, if need be, ‘with learning put lightly, like powder in jam’ ? If you feel you have it in you to lead the way we will try to follow, and perhaps some of the youngest of us may succeed. Wouldn’t this be a triumph worth working for? Try. “Yours very truly, “H.B. DIXON”

THE OWENS COLLEGE, MANCHESTER, May, 1901

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CHAPTER VII. HOW TO SOLVE DIFFERENTIAL EQUATIONS. Table of Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 § 117. Solution by Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 § 118. What is a Differential Equation? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 § 119. Exact Differential Equations of First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 § 120. How to find Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 § 121. The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 § 122. Linear Differential Equations of First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .296 § 123. Equations of First Order and Higher Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 § 124. Clairault’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 § 125. Singular Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 § 126. Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 § 127. Symbols of Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .304 § 128. The Linear Equation of nth Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 § 129. Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 § 130. How to find Particular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 § 131. The Linear Equation with Variable Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 § 132. The Exact Linear Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 § 133. The Integration of Equations with Missing Terms . . . . . . . . . . . . . . . . . . . . . . . . . 319 § 134. Equations of Motion, Chiefly Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 322 § 135. The Velocity of Simultaneous and Dependent Chemical Reactions . . . . . . . . . 330 § 136. Simultaneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 § 137. Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 § 138. What is the Solution of a Partial Differential Equation? . . . . . . . . . . . . . . . . . . 341 § 139. The Solution of Partial Diff. Equations of First Order . . . . . . . . . . . . . . . . . . . . 344 § 140. Partial Differential Equations of nth Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 § 141. Linear Partial Diff. Equations with Constant Coefficients . . . . . . . . . . . . . . . . . 347 § 142. The Particular Integral of Linear Partial Diff. Equations . . . . . . . . . . . . . . . . . . 351 § 143. The Linear Partial Equation with Variable Coefficients . . . . . . . . . . . . . . . . . . . 354 § 144. The Integration of Differential Equations in Series . . . . . . . . . . . . . . . . . . . . . . . . 355 § 145. Harmonic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 Supplemental material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . begins after page 359 The Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-24 The Method of Laplace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-31 Resonant Systems Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-61 Solution to Keplerian Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .S-66 Introduction to Fourier’s Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S-67 Selected sections referenced in chapter VII . . . . . . . . . . . . . . begins after page S-69 § 20. Leibniz’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 § 22. Euler’s Theorem of Homogeneous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 § 23. Successive Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 § 24. Exact Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 § 25. Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .58 § 67. Envelopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .142 § 72. How to Find a Value for the Integration Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 162

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Public domain rationale: J.W. Mellor was a British subject and his works originally published by British publishers. Hence under British copyright law’s principle of “life plus 70 years,” the works of J.W. Mellor passed into the public domain on 1-January-2009.

CHAPTER VII. HOW TO SOLVE DIFFERENTIAL EQUATIONS. THIS chapter may be looked upon as a sequel to that on the integral calculus, but of a more advanced character. The “methods of integration” already described will be found ample for most physico-chemical processes, but chemists are proving every day that more powerful methods will soon have to be brought in. As an illustration, I may refer to the set of differential equations which Geitel encountered in his study of the velocity of hydrolysis of the triglycerides by acetic acid (Journal f¨ ur praktische Chemie [2], 55, 429, 1897). I have previously pointed out that in the effort to find the relations between phenomena, the attempt is made to prove that if a limited number of hypotheses are prevised, the observed facts are a necessary consequence of these assumptions. The modus operandi is as follows: 1. To “anticipate Nature” by means of a “working hypothesis,” which is possibly nothing more than a “convenient, fiction”. “From the practical point of view,” says Professor Bucker (Presidential Address to the B. A. meeting at Glasgow, September, 1901), “it is a matter of secondary importance whether our theories and assumptions are correct, if only they guide us to results in accord with facts. . . . By their aid we can foresee the results of combinations of causes which would otherwise elude us.”

2. Thence to deduce an equation representing the momentary rate of change of the two variables under investigation. 3. Then to integrate the equation so obtained in order to reproduce the “working hypothesis” in a mathematical form suitable for experimental verification (see §§ 18, 69, 88, 89, and elsewhere). So far as we are concerned this is the ultimate object of our integration. By the process of integration we are said to solve the equation.

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For the sake of convenience, any equation containing differentials or differential coefficients will, after this, be called a, differential equation. § 117. The Solution of a Differential Equation by the Separation of the Variables. The different equations hitherto considered have required but little preliminary arrangement before integration. For example, when preparing the equations representing the velocity of a chemical reaction of the general type: dx = kf (t) , dt

(1)

we have invariably collected all the x’s to one side, the t’s, to the other, before proceeding to the integration. This separation of the variables is nearly always attempted before resorting to other artifices for the solution of the differential equation, because the integration is then comparatively simple. The following examples will serve to emphasise these remarks: EXAMPLES. – (1) Integrate the equation, y dx + x dy = 0. Rearrange the terms so that Z Z dx dy dx dy + = 0; or, + = C, x y x y

by multiplying through with 1/xy. Ansr. log x + log y = C. Two or more apparently different answers may be the same. Thus, the solution of the preceding equation may also be written, log xy = log eC , i.e., xy = eC ; or log xy = log C 0 , i.e., xy = C 0 . C and log C 0 are, of course, the arbitrary constants of integration. (2) The equation for the rectilinear motion of a particle under the influence of an attractive force from a fixed point is v Solve. Ansr.

1 2 2v

=

µ x +
2

µ dv = 0. + dx x2

C.

√ √ dy =  y · dx.Ansr. 2 y − tan−1 x = C. 1−a dy dy . Ansr. y = C (a + x) . = a y + dx (4) Solve y − x · dx

(3) Solve 1 + x

(5) In consequence of imperfect insulation, the charge on an electrified body is dissipated at a rate proportional to the magnitude R of the charge. Hence show that if a is a constant depending on the nature of the body, and R, represents the magnitude of the charge when t (time) = 0, E = E0 e−at . Hint. Compound interest law. Integrate by the separation of the variables. Interpret your result. (6) Abegg’s formula for the relation between the dielectric constant (D) of a fluid and temperature θ, is D dD = . − dθ 190

283

Hence show that D = Ceθ/190 , where C is a constant whose value is to be determined from the conditions of the experiment. Put the answer into words. (7) What curves have a slope −x/y to the x-axis? Ansr. The rectangular hyperbolas xy = C. Hint. Set up the proper differential equation and solve. (8) The relation between small changes of pressure and volume of a gas under adiabatic conditions, is γp dx + v dp = 0. Hence show that pv γ = constant. (9) A lecturer discussing the physical properties of substances at very low temperatures, remarked “it appears that the specific heat of a substance decreases with decreasing temperatures at a rate proportional to the specific heat of the substance itself ”. Set up the differential equation to represent this “law” and put your result in a form suitable for experimental verification. (10) Helmholtz equation for the strength of an electric current (C) at the time t, is C=

←−

L dC E − , R R dt

where E represents the electromotive force in a circuit of resistance R and self-induction L. If E, R, L, are constants, show that RC = E(1 − e−Rt/L ) provided C = 0, when t = 0.

A substitution will often enable an equation to be treated by this simple method of solution. EXAMPLE, – Solve (x − y 2 )dx + 2xy dy = 0. Ansr. xey divide by x2 ∴ dx/x + d(v/x) = 0, etc.

2

/x

= C. Hint, put y 2 = v,

If the equation is homogeneous in x and y, that is to say, if the sum of the exponents of the variables in each term is of the same degree, a preliminary substitution of x = ty, or y = tx, according to convenience, will always enable variables to be separated. The rule for the substitution is to treat the differential coefficient which involves the smallest number of terms. dy − 2y = 0. Substitute y = tx, EXAMPLES, – (1) x + y dx Z Z t dt dx 1 ∴ + = C; ∴ + log x = C. x 1−t (1 − t)2

Ansr. (y − x)ex/(x−y) = C. (2) If (y − x)dy + ydx = 0, y = Ce−x/y . (3) If x2 dy − y 2 dx − xy dx = 0, x = ex/y + C. (4) (x2 + y 2 )dx = 2xy dy, x2 − y 2 = Cx.

Non-homogeneous equations in x and y can be converted into the homogeneous form by a suitable substitution. The most general type of a non-homogeneous equation of the first degree is, (ax + by + c) dx + (a0 x + b0 y + c0 ) dy = 0. (2)

284

←−

To convert this into an homogeneous equation, assume that x = v + h and y = w + k, and substitute in the given equation (2). Thus, we obtain {av + bw + (ah + bk + c)} dv + {a0 v + b0 w + (a0 h + b0 k + c0 )} dw = 0. (3) Find h and k so that ah + bk + c = 0; a0 h + b0 k + c0 = 0. (4) b0 c − bc0 ac0 − a0 c ∴ h= 0 ; k = . a b − ab0 a0 b − ab0 Substitute these values of h and k in (3). The resulting equation (av + bw) dv + (a0 v + b0 w) dw = 0 is homogeneous and, therefore, may be solved as just indicated.

(5)

EXAMPLES, – (1) Solve (3y − 7x − 7) dx + (7y − 3x − 3) dy = 0. Ansr. (y − x − 1)2 (y + x + 1)5 = C. Hints. From (2), a = −7, b = 3, c = −7, a0 = −3, b0 = 7, c0 = −3. From (4), h = −1, k = 0. Hence from (3), 3w dv − 7v dv + 7w dw − 3v dv = 0. To solve this homogeneous equation, substitute w = vt, as above, and separate the variables. Z Z Z dv 2dt 5dt 3 − 7t dv = 2 dt; ∴ 7 + + = C. ∴ 7 v t −1 v t−1 t+1 2

5

∴ 7 log v + 2 log (t − 1) + 5 log (t + 1) = C; or v 7 (t − 1) (t + 5) = C.

But x = v + h ∴ v = x + 1; y = w + k ∴ y = w; ∴ t = w/v = y/ (x + 1), etc. (2) If (2y − x − 1) dy + (2x − y + 1) dx = 0; x2 − xy + y 2 + x − y = C.

If in (3),

a : b = a0 : b0 = 1 : m (say), h and k are indeterminate, since (2) then becomes, (ax + by + c) dx + {m (ax + by) + c0 } dy = 0. The denominators in equations (4) also vanish. In this case put z = ax + by and eliminate y, thus, we obtain, dz z+c + = 0, a+b mz + c dx an equation which allows the variables to be separated.

(6)

EXAMPLES, – (1) Solve (2x + 3y − 5) dy + (2x + 3y − 1) dx = 0. Ansr. x + y − 4 log (2x + 3y + 7) = C.

(2) Solve (3y + 2x + 4) dx − (4x + 6y − 1) dy = 0.

Ansr. 9 log {(21y + 14x + 22) /7} − 21 (2y − x) = C.

When the variables cannot be separated in a satisfactory manner, special artifices must be adopted. We shall find it the simplest plan to adopt the routine method of referring each artifice to the particular class of equation which it is best calculated to solve.

285

←−

These special devices are sometimes far neater and quicker processes of solution than the method just described. We shall follow the conventional x and y rather more closely than in the earlier part of this work. The reader will know, by this time, that his x and y’s, his p and v’s and his s and t’s are not to be kept in “water-tight compartments.” It is perhaps necessary to make a few general remarks on the nomenclature. § 118. What is a Differential Equation? We have seen that the straight line, y = mx + b, (1) fulfils two special conditions: 1. It cuts one of the coordinate axes at a distance b from the origin. 2. It makes an angle tan a = m, with the x-axis. By differentiation. dy = m. (2) dx This equation has nothing at all to say about the constant b. That condition has been eliminated. Equation (2), therefore, represents a straight line fulfilling one condition, namely, that it makes an angle tan−1 m with the x-axis. Now substitute (2) in (1), the resulting equation, y=

dy x + b, dx

(3)

in virtue of the constant b, satisfies only one definite condition, (3), therefore, is the equation of any straight line passing through b. Nothing is said about the magnitude of the angle tan−1 m. Differentiate (2). The resulting equation, d 2y = 0, dx2

(4)

represents any straight line whatever. The special conditions imposed by the constants m and b in (1), have been entirely eliminated. Equation (4) is the most general equation of a straight line possible, for it may be applied to any straight line that can be drawn in a plane. Let us now find a physical meaning for the differential equation.

286

In § 7, we have found that the third differential coefficient, d 3 s/dt3 represents “the rate of change of acceleration from moment to moment”. Suppose that the acceleration d 2 s/dt2 , of a moving body does not change or vary in any way. It is apparent that the rate of change of a constant or uniform acceleration must be zero. In mathematical language, this is written, d 3s = 0. (5) dt3 Now integrate this equation once. We obtain d 2s = constant, say, g. dt2

(6)

Equation (6) tells us not only that the acceleration is constant, but it fixes that value to the definite magnitude g ft. per second. But acceleration measures the rate of change of velocity. Integrate (6), we get, ds = gt + C1 . (7) dt From § 72, we have learnt how to find the meaning of C1 . Put t = 0, then dx/dt = C1 . This means that when we begin to reckon the velocity, the body may have been moving with a definite velocity C1 . Let C1 = v0 ft. per second. Of course, if the body started from a position of rest, C1 = 0. Now integrate (7) and find the value of C2 in the result, s = 12 gt2 + v0 t + C2 ,

(8)

by putting t = 0. It is thus apparent that C2 represents the space which the body had traversed when we began to study its motion. Let C2 = s0 ft. The resulting equation s = 21 gt2 + v0 t + s0 ,

(9)

tells us three different things about the moving body at the instant we began to take its motion into consideration. l. It had traversed a distance of s0 ft. To use a sporting phrase, if the body is starting from “scratch,” s0 = 0. 2. The body was moving with a velocity of v0 ft. per second. 3. The velocity was increasing at the uniform rate of g ft. per second. Equation (7) tells us the two latter facts about the moving body; equation (6) only tells us the third fact; equation (5) tells us nothing more than that the acceleration is constant. (5), therefore, is true of the motion of any body moving with a uniform acceleration. EXAMPLE. – If a body falls in the air, experiment shows that the retarding effect of the resisting air is proportional to the square of the velocity of the moving body. Instead of g, therefore, we must write g − βv 2 , where β is the

287

§ 72 is included in this document following the supplement.

variation constant of page 487. For the sake of simplicity, put β = g/a2 and show that v=a

egt/a + e−gt/a a2 gt egt/a − e−gt/a a2 log = log cosh , ; s= gt/a −gt/a g 2 g a e +e

since v = 0 when t = 0, and s = 0 when t = 0.

Similar reasoning holds good from whatever sources we may draw our illustrations. We are, therefore, able to say that a differential equation, freed from constants, is the most general way of expressing a, natural law. Any equation can be freed from its constants by combining it with the various equations obtained by differentiation of the given equation as many times as there are constants. The operation is called elimination. EXAMPLES.–(1) Eliminate the arbitrary constants a and b, from y = ax + bx2 . Differentiate twice and combine the results with the original equation. result, x2

d 2y dy − 2x + 2y = 0, 2 dx dx

is quite free from the arbitrary restrictions imposed in virtue of the presence of the constants a and b in the original equation. (2) Eliminate m from y 2 = 4mx. Ansr. y 2 = 2x dy/dx. (3) Eliminate α and β from y = α cos x + β sin x. Ansr. d 2 y/dx2 + y = 0. (4) Eliminate α and β from y = αeax + βebx . Ansr. d 2 y/dx2 − (a + b)dy/dz + aby = 0. (5) Eliminate k from dx/dt = k(a − x) of § 69. What does the resulting equation mean?

We always assume that every differential equation has been obtained by the elimination of constants from a given equation called the primitive. In practical work we are not so much concerned with the building up of a differential equation by the elimination of constants from the primitive, as with the reverse operation of finding the primitive from which the differential equation has been derived. In other words, we have to find some relation between the variables which will satisfy the differential equation. Given an expression involving x, y, dy/dx, d 2 y/dx2,. . . , to find an equation containing only x, y and constants which can be reconverted into the original equation by the elimination of the constants. This relation between the variables and constants which satisfies the given differential equation is called a general solution, or a complete solution, or a complete integral of the differential

288

←−

equation. A solution obtained by giving particular values to the arbitrary constants of the complete solution is a particular solution. Thus y = mx is a complete solution of y = x dy/dx; p = x tan 45◦ , is a particular solution. A differential equation is ordinary or partial, according as there is one or more than one independent variables present. Ordinary differentia1 equations will be treated first. Equations like (2) and (8) above, are said to be of the first order. For a similar. reason (4) and (6) are of the second order, (5) of the third order. The order of a differential equation, therefore, is fixed by that of the highest differential coefficient it contains. The degree of a differential equation is the highest power of the highest order of differential coeflicient it contains. Thus,  3 d 2y dy +k + µx4 = 0, 2 dx dx

is of the second order and third degree. It is not difficult to show that the complete integral of a differential equation, of the nth order, contains n and only n arbitrary constants. We shall first consider equations of the first order. § 119. Exact Differential Equations of the First Order. The reason many differential equations are so difficult to solve is due to the fact that they have been formed by the elimination of constants as well as by the elision of some common factor from the primitive. Such an equation, therefore, does not actually represent the complete or total differential of the original equation or primitive. The equation is then said to be inexact. On the other hand, an exact differential equation is one that has been obtained by the differentiation of a function of x and y and performing no other operation involving x and y. Easy tests were described in §§ 24, 25, to determine whether any given differential equation is exact or inexact. It was pointed out that the differential equation, M dx + N dy = 0, (1) is the direct result of the differentiation of any function u, provided, ∂M ∂N = . ∂y ∂x

289

(2)

This last result was called the criterion of integrability, because, if an equation satisfies the test, the integration can be readily performed by a direct process. This is not meant to imply that only such equations can be integrated as satisfy the test, for many equations which do not satisfy the test can be solved in other ways. EXAMPLES.–(1) Apply the test to the equations, y dx + x dy = 0 and y dx − x dy = 0. In the former, M = y, N = x; ∴ ∂M/∂y = 1, ∂N/∂x = 1; ∴ ∂M/∂y = ∂N/∂x. The test is, therefore, satisfied and the equation is exact. In the other equation, M = y, N = −x, ∴ ∂M/∂y = 1, ∂N/∂x = −1. This does not satisfy the test. In consequence, the equation cannot be solved by the method for exact differential equations.
(2) Is the
equation, (x + 2y) x dx + x2 − y 2 dy = 0 exact? M = x (x + 2y), N = x2 − y 2 , ∴ ∂M/∂dy = 2x, ∂N/∂x = 2x. The condition is satisfied, the equation is exact.

(3) Show that a2 y + x2 dx + b3 + a2 x dy = 0, is exact. (4) Show that (sin y + y cos x) dx + (sin x + x cos y) dy = 0, is exact.

To integrate an equation which satisfies the criterion of integrability, we must remember that M is the differential coefficient of u with respect to x, y being constant, and N is the differential coefficient of N with respect to y, x being constant. Hence we may integrate M dx on the supposition that y is constant and then treat N dy as if x were a constant. The complete solution of the whole equation is obtained by equating the sum of these two integrals to an undetermined constant. The complete integral is u = C.

(3)


EXAMPLES.–(1) Integrate x (x + 2y) dx + x2 − y 2 dy = 0, from the preceding set of examples. Since the equation is exact, M = x (x + 2y) ; N = x2 − y 2 ; Z Z ∴ M dx = x (x + 2y) dx = 31 x3 + x2 y = Y,

where Y is the integration constant which may, or may not, contain y, because y has here been regarded as a constant. Now the result of differentiating 1 3 3x

+ x2 y = Y,

should be the original equation. On trial, x2 dx + 2xy dx + x2 dy = dY. On comparison with the original equation, it is apparent that dY = y 2 dy; ∴ Y = 13 y 3 + C. Substitute this in the preceding result. The complete solution is, therefore, 1 3 3x

+ x2 y − 31 y 3 = C.

290

To summarise: The method detailed in the example just given may be put into a more practical shape. R To integrate an exact differential equation, first find M dx on the assumption that y is constant and substitute the result in  Z Z Z  ∂ M dx dy = C. (4) N− M dx + ∂y R With the old example, therefore, having found M dx, we may write down at once   Z  ∂ 1 3 2 2 2 2 1 3 x +x y dy = C. x −y − 3x + x y + ∂y 3 Z
x2 − y 2 − x2 dy = C. ∴ 31 x3 + x2 y + And the old result follows directly. If we had wished we could have used  Z Z  Z ∂ M− N dy + N dy dx = C, ∂x

in place of (4). In practice it is often convenient to modify this procedure. If the equation satisfies the criterion of integrability, we can easily pick out terms which make M dx + N dy = 0, and get M dx + Y and N dy + X, where Y cannot contain x and X cannot contain y. Hence if we find M dy and N dx, the functions X and Y will be determined. In the above equation, the only terms containing x and y are 2xy dx + x2 dy, which obviously have been derived from x2 y. Hence integration of these and the omitted terms gives the above result.

(2) Solve x2 − 4xy − y 2 dx + y 2 − 4xy − 2x2 dy = 0. Pick out terms in x and y,

we get − 4xy + 2y 2 dx − 4xy + 2x2 dy = 0. Integrate. ∴ −2x2 y − 2xy 2 = constant. Pick out the omitted terms and integrate for the complete solution. We get, Z Z x2 dx + y 2 dy − 2x2 y − 2xy 2 = x3 − 6x2 y − 6xy 2 + y 3 = constant.
(3) Show that the solution of a2 x + x2 dx + b3 + z 2 x)dy = 0 is a2 xy + b3 y + 13 x3 = C. Use (4).

(4) Solve x2 − y


2

dx − 2xy dy = 0. Ansr.

1 2 3x

− y 2 = C/x. Use (4).

Equations made exact by means of integrating factors. As just pointed out, the reason any differential equation does not satisfy the criterion of exactness, is because the “integrating factor” has been cancelled out during the genesis of the equation from its primitive. If, therefore, the equation M dx + N dy = 0, does not satisfy the criterion of integrability, it will do so when the factor, previously divided out, is restored. Thus, the preceding equation is made exact by multiplying through with the integrating factor µ. Hence, µ (M dx + N dy) = 0, satisfies the criterion of exactness, and the solution can be obtained as described above. 291

§ 120. How to find Integrating Factors. Sometimes integrating factors are so simple that they can be detected by simple inspection. EXAMPLES.–(1) y dx − x dy = 0 is inexact. It becomes exact by multiplication with either x−2 , x−1 y −1 , or y −2 . (2) In (y − x) dy+y dx = 0, the term containing y dx−x dy is not exact, but becomes so when multiplied as in the preceding example. ∴

dy x dy − y dx y ; or log y − = C. − 2 y y x

For the general theorems concerning the properties of integrating factors, the reader must consult some special treatise, say Boole’s A Treatise on Differential Equations, pages 55 et seq., 1865.

We have already established, in § 25, that an integrating factor always exists which will make the equation M dx + N dy = 0, an exact differential. Moreover, there is also an infinite number of such factors, for if the equation is made exact when multiplied by µ, it will amain exact when multiplied by any function of µ. The different integrating factors correspond to the various forms in which the solution of the equation may present itself. For instance, the integrating factor x−1 y −1 , of y dx + x dy = 0, corresponds with the solution log x + log y = C. The factor y −2 corresponds with the solution xy = C 0 . Unfortunately, it is of no assistance to know that every differential equation has an infinite number of integrating factors. No general practical method is known for finding them. Here are a few elementary rules applicable to special cases. Rule I. Since d (xm y n ) = xm−1 y n−1 (my dx + nx dy) , an expression of the type my dx+nx dy, has an integrating factor xm−1 y n−1 ; or, the expression xα y β (my dx + nx dy) = 0, (1) has an integrating factor xm−1−α y n−1−β , or more generally still, xkm−1−α y kn−1−β ,

(2)

where k may have any value whatever. EXAMPLE. –Find an integrating factor of y dx + x dy = 0. Here, α = 0, β = 0, m = 1, n = −1 ∴ y −2 is an integrating factor of the given equation.

292

§ 25 is included in this document following the supplement.

If the expression can be written 0

0

xα y β (my dx + nx dy) + xα y β (m0 y dx + n0 x dy) ,

(3)

the integrating factor can be readily obtained, for 0

0

0

0 0

0

xkm−1−α y kn−1−β ; and xk m −1−α y k n −1−β , are integrating factors of the first and second members respectively. In order that these factors may he identical, km − 1 − α = k 0 m0 − 1 − α0 ; kn − 1 − β = k 0 n0 − 1 − β 0 .

Values of k and k 0 can be obtained to satisfy these two conditions by solving these two equations. Thus, k=

n0 (α − α0 ) − m0 (β − β 0 ) n (α − α0 ) − n (β − β 0 ) 0 ; k = . mn0 − m0 n mn0 − m0 n

(4)

EXAMPLES.–(1) Solve y 3 (y dx − 2x dy) + x4 (2y dx + x dy) = 0. Hints. Show that α = 0, β = 3, m = 1, n = −2, α0 = 4, β 0 = 0, m0 = 2, n0 = 1; ∴ xk−1 y −2k−4 is 0 0 an integrating factor of the first, x2k −5 y k −1 of the second member. Hence, from (4), 0 −3 k = −2, k = 1, ∴ x is an integrating factor of the whole expression. Multiply through and integrate for
2x4 y − y 4 = Cx2
.
(2) Solve y 3 − 2yx2 dx + 2xy 2 − x3 dy. Ansr. x2 y 2 y 2 − x2 = C. Integrating factor deduced after rearranging the equation is xy.

Rule II. If the equation is homogeneous and of the form: M dx+N dy = 0, then (Mx + Ny)−1 an integrating factor. Let the expression M dx + N dy = 0, be of the mth degree and µ. an integrating factor of the nth degree, ∴ µM dx + µN dy = du,

(5)

is of the (m + n)th degree, and the integral u is of the (m + n + 1)th degree. By Euler’s theorem, § 22, ∴ µMx + µNy = (m + n + 1) u.

(6)

Divide (5) by (6), 1 du M dx + N dy = Mx + Ny m+n+1 u The right side of this equation is a complete differential, consequently, the left side is also a complete differential. Therefore, (Mx + Ny)−1 has made M dx + N dy = 0 an exact differential equation.
−1
EXAMPLES.–(1) Show that x3 y − xy 3 is an integrating factor of x2 y + y 3 dx− 2xy 2 dy = 0.
(2) Show that 1/ x2 − nyx + y 2 is an integrating factor of y dy + (x − ny) dx = 0. The method, of course, cannot be used if M x + N y is equal to zero. In this case, we may write y = Cx, a solution.

293

§ 22 is included in this document following the supplement.

Rule III. If the equation is of the form, f1 (x, y) y dx + f2 (x, y) x dy = 0, then (Mx − Ny)−1 is an integrating factor. EXAMPLE. –Solve (1 + xy) y dx+(1 − xy) x dy = 0. Hint. Show that the integrating R factor is 1/2xy. Divide out 12 . ∴ M dx = 1/xy + log x. Ansr. x = Ce−1/xy If M x − N y = 0, the method fails and xy = C is then a solution of the equation. E.g., (1 + xy) y dx + (1 + xy) x dy = 0. 1 N



∂M ∂y

∂N ∂x

R



is a function of x only, e f (x)dx is an inte R ∂M f (y)dy grating factor. Or, if M1 ∂N is an integrating − = f (y), then e ∂x ∂y factor. These are important results. Rule IV. If

− 


EXAMPLES.–(1) Solve x2 + y 2 dx − 2xy dy = 0. Ansr. x2 − y 2 = Cx. Hint. Show f (x) = −2/x. The integrating factor is, therefore, e−

R

2dx/x

2

= elog 1/x = 1/x2 .

(Why?) Prove that this
is an integrating factor,
and solve as in the preceding section. (2) Solve y 4 + 2y dx + xy 3 + 2y 4 − 4x dy = 0. Ansr. xy 3 y 4 + 2x = Cy2. (3) We may prove the rule for a special case in the following manner. The steps will serve to recall some of the principles established in some earlier chapters. dy + P y = Q, dx

Let

(7)

where P and Q are either constant or functions of x. Let µ be an integrating factor which makes dy + (P y − Q) dx = 0, (8) an exact differential. ∴ µ dy + µ (P y − Q) dx ≡ N dy + M dx. ∂N ∂µ ∂M ∂µ ∴ = ; = (P y − q) + P µ. ∂x ∂x ∂y ∂y ∂µ ∂µ = (P y − Q) + P µ. ∴ ∂x ∂dx ∂µ ∂µ dx = (P y − Q) dx + P µ dx; ∴ ∂x ∂y ∂µ = − dy + P µ dx. ∂y ∂µ ∂µ ∴ dx + dy = dµ = P µ dx ∂x ∂y Z 1 dµ ∴ P = ; P dx = log µ. µ dx and since loge e = 1. Z

 P dx log e = log µ;

µ=e

R

P dx

.

This result will be employed in dealing with linear equations, § 122.

294

(9)

←−

§ 121. The First Law of Thermodynamics. According to the discussion at the end of the first chapter, one way of stating the first law of thermodynamics is as follows: dQ = dU + dW, which means that when a quantity of heat, dQ, is added to a substance, one part of the heat is spent in changing the internal energy, dU, of the substance and another part, dW , is spent in doing work against external forces. In the special case, when that work is expansion against atmospheric pressure, dW = p dv, as shown in § 91. See (11), page 524. We know that the condition of a substance is completely defined by any two of the three variables p, v, θ, because when any two of these three variables is known, the third can be deduced from the relation pv = Rθ. Hence it is assumed that the internal energy of the substance is completely defined when any two of these variables are known. Now let the substance pass from any state A to another state S (Fig. 111). The internal energy of the substance in the state B completely determined by the coordinates of that point, because U is quite independent of the nature of the transformation from the state A to the state B. It makes no difference to the magnitude of U whether that path has been via AP E or AQB. In this case U is said to be a singlevalued function by the coordinates of the point corresponding to any given state. In other words, dU is a complete differential. Figure 111: Hence dU =

∂U ∂U dx + dy, ∂x ∂y

is an exact differential equation, where x and y represent any pair of the variables p, v, θ. On the other hand, the external work done durmg the transformation from the one state to another, depends not only on the initial and final states of the substance, but also on the nature of the path described in passing from the state A to the state B. For example, the substance may perform the work represented by the area AQBB 0 A0 or by the area AP BB 0 A0 , in its passage from the

295

state A to the state B. In fact the total work done in the passage from A to B and back again, is represented by the area AP BQ (page 183). In order to know the work done during the passage from the state A. to the state B, it is not only necessary to know the initial and final states of the substance as defined by the coordinates of the points A and B, but we must know the nature of the path from the one state to the other. Similarly, the quantity of heat supplied to the body in passing from one state to the other, not only depends on the initial and final states of the substance but also on the nature of the transformation. All this is implied when it is said that “dW and dQ are not perfect differentials.” Although we can write ∂2U ∂2U = , ∂x∂y ∂y∂x we must put, in the case of W or Q, ∂2Q ∂2Q ≶ ; or ∂x∂y ∂y∂x

∂2Q ∂2Q = . ∂x∂y ∂y∂x

Therefore the partial differentiation of x with respect to y, furnishes a complete differential equation only when we multiply through with the integrating factor µ, so that µ dQ = µ

∂Q ∂Q dx + µ dy, ∂x ∂y

where x and y may represent any pair of the variables p, v, θ. The integrating factor is proved in thermodynamics to be equivalent to the so-called Carnot’s function (see Preston’s Theory of Heat). To indicate that dW and dQ are not perfect differentials, some writers superscribe a comma to the top right-hand corner of the differential sign. The above equation would then be written, d 0 Q = dU + d 0 W. § 122. Linear Differential Equations of the First Order. A linear differential equation of the first order involves only the first power of the dependent variable y and of its first differential coefficients. The general type is, dy + P y = Q, (1) dx where P and Q may be functions of x, or constants.

296

R

We have just proved that e R

e

P dx

P dx

is an integrating factor of (1), therefore R

(dy + P y dy) = e

P dx

Q dx,

is an exact differential equation. The general solution is, Z R R P dx ye = e P dx Q dx + C.

(2)

The linear equation is one of the most important in applied mathematics. In particular cases the integrating factor may assume a veryR simple form. In the following examples, remember that elog x = x, ∴ if P dx = log x, R e P dx = x.
EXAMPLES.–(1) Solve 1 + x2 dy = (m + xy) dx. Reduce to the form (1) and we obtain dy m x y= . − 2 dx 1 + x 1 + x2 Z Z p
x dx 1 ∴ P dx = − = − log 1 + x2 = − log (1 + x2 ). 2 1+x 2

Remembering log 1 = 0, log e = 1, the integrating factor is evidently, R p R 1 . log e P dx = log 1 − log 1 + x2 , or e P dx = √ 1 + x2

Multiply the original equation with this integrating factor, and solve the resulting p exact equation as § 119, (4), or, better still, by (2) above. The solution: y = mx+C (1 + x2 ), follows at once. (2) Ohm’s law for a variable current flowing in a circuit with a coefficient of selfinduction L (henries), a resistance R (ohms), and a current of C (amperes) and an electromotive force E (volts), is given by the equation, E = RC + L

dC . dt

This equation has the standard linear form (1). If E is constant, show that the solution is, C = E/R + Be−Rt/L , where B is the arbitrary constant of integration (page 159). Show that C approximates to E/B after the current has been flowing some time (t). Hint for solution. Integrating factor is eRt/L . (3) The equation of motion of a particle subject to a resistance varying directly as the velocity and as some force which is a given function of the time, is dv/dt + kv = f (t) . R Show that v = Ce−kt + e−kt ekt f (t) dt. If the force is gravitational, say g, v = Ce−kt + g/k

(4) Solve x dy + y dx = x3 dx. Integrating factor= x. Ansr. y = 41 x3 + C/x.

Many equations may be transformed into the linear type of equation, by a change in the variable. Thus, in the so-called Bernoulli’s equation, dy/dx + P y = Qy n . 297

(3)

Divide by y n , multiply by (1 − n) and substitute y n−1 = v, in the result. Thus, 1 − n dy + (1 − n) P y 1−n = (1 − n) Q, y n dx and dy/dx + (1 − n) P v = Q (1 − n), which is linear in v. Hence, the solution is Z R R (1−n) P dx ve = (1 − n) Qe(1−n) P dx dx + C. Z R R 1−n (1−n) P dx ∴ y e = (1 − n) Qe(1−n) P dx dx + C. EXAMPLES.–(1) Solve dy/dx + y/x = y 2 . Treat as above, substituting v = 1/y. The R dx/x integration factor is e = elog x = 1/x. Ansr. Cxy − xy log x = 1.

(2) Solve dy/dx +R x sin 2y = x3 cos2 y. Divide by cos2 y. Put tan y = v. The
2 2 2 integration factor is e 2x dx , i.e., ex . Ansr. ex tan y − 12 ex x2 − 1 = C. Hint to R 2 x2 solve = x3 ex dx + C. Put x2 = z, ∴ 2x dx = dz, and this integral becomes R ve 1 1 z z ze dz, or 2 e (z − 1), etc. 2 (3) Here is an instructive differential equation, which Harcourt and Esson encountered during their work on chemical dynamics in 1866. K K 1 dy + − = 0. 2 y dx y x I shall give a method of solution in full, so as to revise some preceding work. The equation has the same form as Bernoulli’s. Therefore, substitute 1 dv 1 dy v = ; i.e., =− 2 . y dx y dx K dv − Kv + = 0, ∴ dx x an equation linear in v. The integrating factor is e

R

P dx

therefore, from (2) From § 108,

, or, e−Kx ; Q, in (2), = −K/x, R −Kx ve−Kx = − K dx + C. x e

( ) 2 3 1 (Kx) (Kx) 1 − (Kx) + ve−Kx = −K − + . . . dx + C. x 1·2 1·2·3  Z  dx K 2 x dx K 3 x2 dx ∴ ve−Kx = −K − K dx + − + . . . + C. x 1·2 1·2·3 Z

But v = 1/y. Multiply through with yeKx , and integrate. ( ) 2 3 (Kx) (Kx) 1 = KeKx C1 − log x + Kx − + − . . . y. 1·2 1·2·3 We shall require this result on page 333.

§ 123. Differential Equations of the First Order and of the First or Higher Degree.– Solution by Differentiation. Case i. The equation can be split up into factors. If the differential equation can be resolved into n factors of the first degree, equate each factor to zero and solve each of the n equations separately. The n solutions may be left either distinct, or combined into one.

298

EXAMPLES.–(1) Solve x



dy dx

2

= y. Resolve into factors of the first degree, dy =± dx

r

y . x

Separate the variables and integrate, √ √ √ x ± y = ± C, which, on rationalisation, becomes 2

(x − y) − 2C (x + y) + C 2 = 0 Geometrically this equation represents a system of parabolic curves each of which touches the axis at a distance C from the origin. The separate equations of the above solution merely represent different branches of the same parabola.  2
dy dy (2) Solve xy dx − xy = 0. Ansr. xy = C, or x2 − y 2 = C. Hint. − x2 − y 2 dx

dy . Factors (xp + y) (yp − x), where p ≡ dx  2 dy dy (3) Solve dx − 7 dx + 12 = 0. Ansr. y = 4x + C, or 3x + C.

Case ii. The equation cannot be resolved into factors, but it can be solved for x, y, dy/dx, or y/x. An equation which cannot be resolved into factors, can often be expressed in terms of x, y, dy/dx, or y/x, according to circumstances. The differential coeffcient of the one variable with respect to the other may be then obtained by solving for dy/dx and using the result to eliminate dy/dx from the given equation. 2

EXAMPLES. Solve dy/dx + 2xy = x2 + y 2 . Since (x − y) = x2 − 2xy + y 2 , p y = x + dy/dx   ,  12 dy dy 1 d2 y . Differentiate = 1 + dx 2 dx2 dx Separate the variables x and p, where p ≡ dy/dx, and solve for dy/dx, r √ p−1 C + e2x 1 dy . = x = log √ + log C; 2 p+1 dx C − e2x

∴ Ansr. y = x + C + e2x / C − e2x .
(2) Solve x (dy/dx)2 − 2y (dy/dx) + ax = 0. Ansr. y = 21 C x2 + a/C . Hint. Substitute for p. Solve for y and differentiate. Substitute p dx for dy, and clear of fractions. The variables p and x can be separated. Integrate. p = xC. Substitute in the given equation for the answer. 2 (3) Solve x (dy/dx) + 2y (dy/dx) − y = 0. Ansr. (2x + C). Hint. Solve for x. Differentiate and substitute dy/p for dx, and proceed as in example (2). yp = C, etc.

Case iii. The equation cannot be resolved into factors, x or y is absent. If x is absent solve for dy/dx or y according to convenience; if y is absent, solve for dx/dy or x. Differentiate the result with respect to the absent letter if necessary and solve in the regular way.

299

See page S-13 of the supplement for more details on the case ii method. KH

←−

2

EXAMPLES.–(1) Solve (dy/dx) + x (dy/dx) + 1 = 0. For the sake of greater ease, substitute p for dx/dy. The given equation thus reduces to x = p + 1/p.

(1)

Differentiate with regard to the absent letter y, thus,
p = 1 − 1/p2 dp/dy; or dy/dp = 1/p + 1/p2 .

(2)

Combining (1) and (2), we get the required solution. (2) Solve dy/dx = y + 1/y. Ansr. y 2 = Ce2x − 1. (3) Solve dy/dx = x + 1/x. Ansr. y = 21 x2 + log x + C.

§ 124. Clairaut’s Equation. The general type of this equation is,   dy dy y=x ; (1) +f dx dx or, writing p = dy/dx, for the sake of convenience, y = px + f (p) .

(2)

Many equations of the first degree in x and y can be reduced to this form by a more or less obvious transformation of the variables, and solved in the following way:– Differentiate (2) with respect to x, and equate the result to zero. p=p+x Hence either If in the former,

dp dp + f 0 (p) ; or, dx dx

{x + f 0 (p)}

dp = 0. dx

dp = 0; or, x + f (p) = 0. dx dp/dx = 0;

∴ p = C,

where C is an arbitrary constant. Hence, dy = C dx;

or, y = Cx + f (C) ,

is a solution of the given equation. Again, p in x + f (p) may be a solution of the given equation. To find p, eliminate p between y = px + f (p) , or, x + f 0 (p) = 0. The resulting equation between x and y also satisfies the given equation. There are thus two classes of solutions to Clairaut’s equation. EXAMPLES. –Find both solutions. in the following equations:– (1) y = px + p2 . Ansr. Cx + C 2 = y and x2 + 4y = 0. √ √ (2) (y − px) (p − 1). Ansr. (y = Cx) (C − 1) = C; y + x = 1. Read over § 67.

300

§ 67 is included in this document following the supplement.

←−

§ 125. Singular Solutions. Clairaut’s equation introduces us to a new idea. Hitherto we have assumed that whenever a function of x and y satisfies an equation, that function plus an arbitrary constant, represents the complete or general solution. We now find that functions of x and y can sometimes be found to satisfy the given equation, which, unlike the particular solution, are not included in the general solution. This function must be considered a solution, because it satisfies the given equation. But the existence of such a solution is quite an accidental property confined to special equations, hence their cognomen, singular solutions. To take the simple illustration of page 149, a y = px + . p

(1)

Remembering that y has been written for dy/dx, differentiate with respect to x, we get, on rearranging terms,   a dp x− 2 = 0. p dx where either

x−

a = 0; or, p2

dp = 0. dx

If the latter,

a (2) p = C; or, y = Cx + . C p If the former, p = a/x, which gives,.when substituted in (1), the solution, y 2 = 4ax. (3) This solution is not included in the general solution, but yet it satisfies the given equation. (3) is the singular solution of (1). Equation (2), the complete solution of (1), has been shown to represent a system of straight lines which differ only in the value of the arbitrary constant C; equation (3), containing no arbitrary constant, is an equation to the common parabola. A point moving on this parabola has, at any instant, the same value of dy/dx as if it were moving on the tangent of the parabola, or on one of the straight lines of equation (2). The singular solution of a differential equation is geometrically equivalent to the envelope of the family of curves represented by the general solution. The singular solution is distinguished from the particular solution, in that the latter is contained in the general solution, the former is not. Again referring to Fig. 78, it will be noticed that for any point on the envelope, there are two equal values of p or dy/dx, one for the parabola, one for the straight line.

301

In order that the quadratic ax2 + bx + c = 0, may have equal roots, it is necessary (page 388) that b2 = 4ac;

or, b2 − 4ac = 0.

(4)

This relation is called the discriminant. From (1), since y = px + a/p;

∴ xp2 − yp + a = 0.

(5)

In order that equation (5) may have equal roots, y 2 = 4ax, as in (4). This relation is the locus of all points for which two values of C become equal, hence it is called the p -discriminant of (1). In the same way if C be regarded as variable in the general solution (2), y = Cx + a/C; or, xC 2 − yC + a = 0. The condition for equal roots, is that y 2 = 4ax. which is the locus of all points for which the value of C is the same. It is called the C -discriminant. Before applying these ideas to special cases, we may note that the envelope locus may be a single curve (Fig. 78) or several (Fig. 79). For an exhaustive discussion of the properties of these discriminant relations l must refer the reader to the numerous textbooks on the subject, or to Cayley, Messenger of Mathematics, 2, 6, 1879. To summarise: 1. The envelope locus satisfies the original equation but is not included in the general solution (see xx0 , Fig. 112).

Figure 112: –Nodal and Tac Loci 2. The tac locus is the locus passing through the several points where two non-consecutive members of a family of curves touch. Such a locus is represented by the line AB (Fig. 79), P Q (Fig. 112). The tac locus does not satisfy the original equation, it appears in the p-discriminant, but not in the C-discriminant. 302

For expanded discussion of discriminants, see supplement page S-17. KH

3. The node locus is the locus passing through the different points where each curve of a given family crosses itself (the point of intersection – node – may be double, triple, etc.). The node locus does not satisfy the original equation, it appears in the C-discriminant but not in the pdiscriminant. RS (Fig. 112) is a nodal locus passing through the nodes A,. . . , B,. . . , C,. . . , M. 4. The cusp locus passes through all the cusps (page 186) formed by the members of a family of curves. The cusp locus does not satisfy the original equation, it appears in the p- and in the C-discriminants. It Figure 113: –Cusp Locus is the line Ox in Fig. 118. Sometimes the nodal or cusp loci coincides with the envelope locus.* EXAMPLES.–Find the singular solutions and the nature of the other loci in the following equations: (1) xp2 − 2yp + ax = 0. 2 2 For equal roots y = ax . This satisfies the original equation and is not included in the general solution: x2 − 2Cy + aC 2 = 0. y 2 = ax2 is thus the singular solution. (2) 4xp2 = (3x − a)2 . General solution: (y + C)2 = x (x − a)2 . 2 For equal roots in p, 4x (8x − a) = 0, or x(3x − a)2 = 0 (p-discriminant). For equal roots in C, differentiate the general solution with respect to C. Therefore 2 (x + C) dx/dC = 0, or C = −x. ∴ x (x − a) = 0 (C-discriminant) is the condition to be fulfilled when the C-discriminant has equal roots. x = 0 is common to the two discriminants and satisfies the original equation (singular solution); x = a satisfies the C-discriminant but not the p-discriminant and, since it is not a solution of the original equation, x = a represents the node locus; x = 13 a satisfies the p- but not the C-discriminant nor the original equation (tac locus). (3) p2 + 2xp − y = 0.
2
3 3 General solution: 2x + 3xy + C = 4 x2 + y ; p-discriminant: x2 + y = 0; C
3 discriminant: x2 + y = 0. The original equation is not satisfied by
either of these equations and, therefore, there is no singular solution. Since x2 + y appears in both discriminants, it represents a cusp locus.
(4) Show that the complete solution of the equation, y 2 p2 + 1 = a2 , is 2 y 2 + (x − C) = a2 ; that there are two singular solutions, y = ±a; that there is a tac locus on the x-axis for y = 0 (Fig. 79, see also § 138). * The second part of van der Waals’ The Continuity of the Gaseous and Liquid States of Aggregation – Binary Mixtures (1900) has some examples of the preceding “mathematics.”

303

§ 126. Trajectories. This section will serve as an exercise on some preceding work. A: trajectory is a curve which cuts another system of curves at a constant angle. If this angle is 90◦ the curve is an orthogonal trajectory. EXAMPLES. (1) Let xy = C be a system of rectangular hyperbolas, to find the orthogonal trajectory, first eliminate C by differentiation with respect to x, thus we obtain, x dy/dx + y = 0
If two curves are at right angles 21 π = 90◦ , then from (17), § 32, 12 π = (α0 − α), where α, α0 are the angles made by tangents to the curves at the point of intersection with the x-axis. But by the same formula,   1 tan ± π = (tan α0 − tan α) / (1 + tan α tan α0 ) . 2 Now tan ± 21 π = ∞ and 1/∞ = 0, ∴ tan α = − cot α0 ; or, dy/dx = −dx/dy.* The differential equation of the one family is obtained from that of the other by substituting dy/dx for −dx/dy. Hence the equation to the orthogonal trajectory of the system of rectangular hyperbolas is, x dx + y dy = 0, or x2 − y 2 = C, a system of rectangular hyperbolas whose axes coincide with the asymptotes of the given system. (2) For polar coordinates show that we must substitute −dr/ (r · dθ) for r · dθ/dr. (3) Find the orthogonal trajectories of the system of parabolas y 2 = 4ax. Ansr. Ellipses, 2x2 + y 2 = C 2 . (4) Show that the orthogonal trajectories of the equipotential curves, 1/r−1/r0 = C, are the magnetic curves cos θ + cos θ0 = C.

§ 127. Symbols of Operation. It will be found convenient to denote the symbol of the operation “dy/dx” by the letter “D”. If we assume that the infinitesimal increments of the independent variable dx have the same magnitude, whatever be the value of x, we can suppose D to have a constant value. Thus d d2 d3 , , ,...; dx dx2 dx3 dy d2 y d3 y , , ,... Dy, D 2y, D 3y, . . . , stand for dx dx2 dx3 D, D 2 , D 3, . . . , stand for

The operations denoted by the symbols D, D 2 ,. . . , satisfy the elementary rules of algebra except that they are not commutative†with regard to the variables. For example, * No doubt the reader sees that in (18), § 12, dx/dy is the cotangent of the angle whose tangent is dy/dx. † The so-called fundamental laws of algebra are: I. The law of association: The number of things in any group is independent of the order. II. The commutative law: (a) Addition. The number of things in any number of groups is independent of the order. (b) Multiplication. The product of two numbers is independent of the

304

(17) of § 32 establishes that slopes of m and −1/m are normal to each other. KH

D (u + v + . . . ) = Du + dv + . . . , (distributive law). D (Cu) = CD (u) , (commutative law), where C is a constant. We cannot write D (xy) = D (yx). But, Dm Dn u = Dm+n u

(index law),

is true when m and n are positive integers. If Du = v; u = D−1 v; ∴ v = D · D−1 v;

1 v; D or, D · D = 1; or, u =

that is to say, by operating with D upon D−1 v, we annul the effect of the D−1 operator. 1 3 It is necessary to remember later on, that if Dx = 1, D12 = 21 x2 ; D13 = 2·3 x ,... In this notation, the equation dy d2 y − (α + β) + αβy = 0, 2 dx dx is written, 

D2 − (α + β) D + αβ y = 0; or,

(D − α) (D − β) y = 0.

Now replace D with the original symbol, and operate on one factor with y, Thus,       d d d dy −α − β y = 0; − αy − βy = 0. dx dx dx dx By operating on the second factor with the first, we get the original equation back again.

§ 128. The Linear Equation of the nth Order. (General Remarks.) As a general rule the higher orders of differential equations are more difficult of solution than equations of the first order. As with the latter, the more expeditious mode of treatment will be to refer the given equation to a set of standard cases having certain distinguishing characters. By far the most important class is the linear equation. A linear equation of the nth order is one in which the dependent variable and its n derivatives are all of the first degree and are not multiplied together. The typical form in which it appears is dn−1 y dn y + X + · · · + Xn y = X. 1 dxn dxn−1

(1)

order. III. The distributive law: (a) Multiplication. The multiplier may be distributed over each term of the multiplicand, e.g., m(a + b) = ma + mb. (b) Division. (a + b)/m = a/m + b/m. IV. The index law: (a) Multiplication. am an = am+n . (5) Division. am /an = am−n .

305

Or, in symbolic notation, D n y + X1 D n−1 y + · · · + Xn y = X, where X, X1 , . . . , Xn , are either constant magnitudes, or functions of the independent variable x. If the coefficient of the highest derivative be other than unity, the other terms of the equation can be divided by this coefficient. The equation will thus assume the typical form (1). We have studied the linear equation of the first order in § 123. For the sake of fixing our ideas, the equation dy d2 y + P + Qy = r, (2) dx2 dx of the second order, will be taken as typical of the class. P , Q, R have the meaning above attached to X1 , X2 , X. The general solution of the linear equation is made my of two parts. 1. The complementary function which is the most general solution of the left-hand side of equation (2) equated to zero, or, d2 y dy + P + Qy = 0. dx2 dx

(3)

The complementary function involves two arbitrary constants. 2. The particular integral which is any solution of the original equation (2), the simpler the better. In particular cases when the right-hand side is zero, the particular integral does not occur. To show that the general solution of (2) contains a general solution of (3). Assume that the complete solution of (2) may be written, y = u + v,

(4)

where v is any function of x which satisfies (2), that is to say, v is the particular integral* of (2), u is the general solution of (3), to be determined. Substitute (4) in (2).

But therefore,

d2 u du d2 v dv +P + Qu + 2 + P + Qv = R. 2 dx dx dx dx dv d2 v +P + Qv = R; 2 dx dx d2 u du + P + Qu = 0. dx2 dx

Therefore, u must satisfy (3). Given a particular solution* of the linear equation, to find the * Not to be confused with the particular solution of page 289.

306

complete solution. Let y = v be a particular solution of the following equation, d2 y dy +P + Qy = 0, 2 dx dx where P and Q are functions of x. Substitute y = uv,   dv du d2 u + Pv = 0. v 2+ 2 dx dx dx This equation is of the first order and linear with du/dx as the dependent variable. Put du/dx = z and   dv dz dv dz + 2 + P v z = 0; +2 + P dx = 0; v dx dx z v Z R du 1 du + 2 log v + P dx = 0; or, = C1 2 e− P dx . log dx dx v Z Z R R 1 − P dx 1 − P dx e + C2 ; or, C1 v e + C2 v, ∴ u = C1 2 v v2 where C1 and C2 are arbitrary constants. EXAMPLES.–(1) If y = eax is a particular solution of d2 y/dx2 = a2 y, show that the complete solution is y = C1 eax + C2 e−ax .
(2) If y = x is a particular solution of 1 − x2 d2 y/dx2 − x dy/dx + y = 0, the √ complete solution is y = C1 1 − x2 + C2 x.

If a particular solution of the linear equation is known, the order of the equation can be lowered by unity. This follows directly from the preceding result. If y = v is a known solution, then, if y = tv be substituted in the first member of the equation, the coefficient of t in the result, will be the same as if t were constant and therefore zero. t being absent, the result will be a linear equation in t but of an order less by unity than that of the given equation. It follows directly, that if n particular solutions of the equation are known, the order of the equation can be reduced n times. For the description of a machine designed for solving (3), see Proceedings of the Royal Society, 24, 269, 1876 (Lord Kelvin). § 129. The Linear Equation with Constant Coefficients. The integration of these equations obviously resolves itself into finding the complementary function and the particular integral. First, when the second member is zero, in other words, to find the complementary function of any linear equation with constant coefficients. The typical equation is, d2 y dy + P + Qy = 0, (1) dx2 dx 307

See page S-21 of the supplement for the derivation of this equation. KH

where P and Q are constants. The particular integral does not appear in the solution. R If the equation were of the first order, its solution would be, y = Ce m dx . On substituting emx for y in (1), we obtain
m2 + P m + Q emx = 0, m2 + P m + Q = 0.

provided

(2)

This equation is called the auxillary equation. If m1 be one value of m which satisfies (1), then y = em1 x , is an integral of (1). But we must go further. Case 1. When the auxillary equation has two unequal roots, say m1 , and m2 the general solution of (1) may be written down without any further trouble. u = C1 em1
x + C2 em2 x . (3)

EXAMPLES.–(1) Solve D2 + 14D − 32 y = 0. Assume y = Cemx is a solution. The auxillary becomes, m2 + 14m − 32 = 0. The roots are m = 2, or m = 16. The required solution is, therefore, y = C1 e2x + C2 e−16x . (2) Solve d2 y/dx2 − m2 y = 0. Ansr. y = C1 emx + C2 e−mx (see page 319). (3) Show that y = C1 e3x + C2 ex is a complete solution of 2

2

d y/dx + 4 dy/dx + 3y = 0.

←−

Case 2. When the two roots of the auxillary are equal. if m1 = m2 in (3), it is no good putting (C1 + C2 ) em1 x as the solution, because C1 +C2 , is really one constant. The solution would then contain one arbitrary constant less than is required for the general solution. To find the other particular integral, it is usual to put m1 = m2 + h, where h is some finite quantity which will ultimately be made zero. With this proviso, we write the solution, y = lim C1 em1 x + C2 e(m2 +h)x . h→0
y = lim em1 x C1 + C2 ehx .

Hence,

h→0

Now expand ehx by Maclaurin’s theorem (page 230). 
∴ y = lim em1 x C1 + C2 1 + hx + 21 h2 x2 + . . . ; h→0 
= lim em1 x C1 + C2 + C2 hx + 2!1 h2 x2 + . . . ; h→0
= lim em1 x A + Bx + 2!1 C2 h2 x2 + C2 R ,

See page S-22 of the supplement for an alternative development of this general solution. KH

h→0

where R denotes the remaining terms of the expansion of ehx , A = C1 + C2 , B = C2 h. Therefore, at the limit, y = em1 x (A + Bx) .

(4)

For the sake of uniformity, we shall still write the arbitrary integration constants C1 , C2 , C3 , . . .

308

For an equation of a still higher degree, the preceding result may be
written, y = em1 x C1 + C2 x + C3 x2 + · · · + Cr−2 xr−1 . (5) where r denotes the number of equal roots.

EXAMPLES.–(1) Solve d3 y/dx3 − d2 y/dx2 − dy/dx + u = 0. Assume y = emx . The auxillary equation is m3 − m2 − m + 1 = 0. The roots are 1, 1, −1. Hence the general solution can be written down at sight: y = C1 e−x + (C2 + C − 3x) ex .
(2) Solve D3 − 3D2 + 4 y = 0. Ansr. e2x (C1 + C2 x) + C3 e−x .

Case 3. When the auxillary equation has imaginary roots, all unequal. Remembering that imaginary roots are always found in pairs in equations with real coefficients (page 386), let the two imaginary roots be m1 = α + ιβ; and m2 = α − ιβ.

Instead of substituting y = emx in (3), we substitute these values of m in (3) and get y = C1 e(α+ιβ)x + C2 e(α−ιβ)x ;
= eαx C1 eιβx + C2 e−ιβx ; = eαx C1 (cos βx + ι sin βx) + C2 (cos βx − ι sin βx) .

(6)

(See the chapter on “Hyperbolic Functions.”) Separate the real and imaginary parts, as in Ex. 3, p. 280, y = eαx (C1 + C2 ) cos βx + ι (C1 − C2 ) sin βx; if we put C1 + C2 = A, ι (C1 − C2 ) = B.

y = eαx (A cos βx + B sin βx) .

(7)

In order that the constants A and B in (7) may be real, the constants C1 and C2 must include the imaginary parts. EXAMPLES.–(1) Show from (6) that y = (cosh αx + sinh αx) (A1 cos βx + B2 sin βx) . (Exercise on Chapter VI.) √ (2) Integrate d2 y/dx2 + dy/dx + y = 0. The roots are α = − 21 and β = 12 3; √ √
∴ y = e−x/2 A cos 12 3x + B sin 21 3x . (3) The equation of a point vibrating under the influence of a periodic force, is, t d2 x + α2 x = a cos 2π . dt2 T Find the complementary function. The roots are ±ια. From (7) x = A cos αt + B sin αt.

3

2

(4) If D − D + D − 1 y = 0, y = C1 cos x + C2 sin x + Cex .


Case 4. When some of the imaginary roots of the auxillary equation are equal. If a pair of the imaginary roots are repeated, 309

See S-66 of the supplement for how case 3 arises in the equations for the orbits of planets. KH

we may proceed as in Case 2, since, when m1 = m2 , C1 em1 x + C2 eem2 x is replaced by (A + Bx) em1 x ; similarly, when m3 = m4 , C3 em3 x + C3 em4 x may be replace by (C + Dx) em3 x . If, therefore, m1 = m2 = α + ιβ; and m3 = m4 = α − ιβ, the solution y = (C1 + C2 x) e(α+ιβ)x + (C3 + C4 x) e(α−ιβ)x , y = eαx (A + Bx) cos βx + (C + Dx) sin βx.

becomes

(8)


EXAMPLES.–(1) Solve D4 − 12D3 + 62D2 − 156D + 169 y = 0. Given the roots of the auxillary: 3 + 2ι, 3 + 2ι, 3 − 2ι, 3 − 2ι. Hence y = e3x {(C1 + C2 x) sin 2x + (C3 + C4 x) cos 2x} . (2) If D2 + 1


2

(D − 1)2 y = 0, (A + Bx) sin x + (C + Dx) cos x + (E + F x) ex .

Second, when the second member is not zero, that is to say to find both the complementary function and the particular integral. The general equation is, dy d2 y +P + Qy = R, (9) 2 dx dx where P and Q are constant, R is a function of x. We have just shown how to find one part of the complete solution of the linear equation with constant coefficients, namely, by putting R, in (9), equal to zero. The remaining problem is to find a particular integral of this equation. The more useful processes will be described in the next section. In the symbolic notation, (9) may be written, f (D) y = R

(10)

The particular integral is, therefore, y = f (D)−1 R; or y =

R . f (D)

(11)

The right-hand side of either of equations (11), will be found to give a satisfactory value for the particular integral in question. Since the complementary function contains all the constants necessary for the complete solution of the differential equation, it follows that no integration constant must be appended to the particular integral. § 130. How to find Particular Integrals. It will be found quickest to proceed by rule: Case 1 (General). When the operator f (D)−1 can be resolved into factors.

310

We have seen that the linear differential equation of the first order, is solved by

dy/dx − ay = R; or y = R/ (D − a) , Z ax e−ax R dx. y=e

(1) (2)

The term Ceax in the solution of (1), belongs to the complementary function. Suppose that in a linear equation of a higher order, say, d2 y/dx2 − 5 dy/dx − 6y = R,

the operator f (D)−1 can be factorised. The complementary function is written down at sight from,
D 2 − 5D + 6 y = 0; or (D − 3) (D − 2) y = 0. y = C1 e3x + C2 e2x .

namely,

(3)

The particular integral is   1 1 1 R; R= − y1 = (D − 3) (D − 2) D−3 D−2 Z Z −3x 2x 3x e−2x R dx, e R dx − e =e

(4)

from (2). The general solution is the sum of (3) and (4), Z Z −3x 2x 3x 2x 3x e−2x R dx. e R dx − e ∴ y = C1 e + C2 e + e

EXAMPLES.–(1) In the preceding illustration, put R = e4x and show that the general solution is, C1 e3x + C2 e
2x + 21 e4x . (2) If D2 − 4D + 3 y = 2e3x , y = C1 ex + C2 e3x + xe3x .

Case 2 (General). When the operator f (D)−1 can be resolved into partial fractions with constant numerators. The way to proceed in this case is illustrated in the first example below.

EXAMPLES.–(1) Solve d2 y/dx2 − 3 dy/dx + 2y = e3x . In symbolic notation this will appear in the form, (D − 1) (D − 2) y = e3x. The complementary function is y = C1 ex + C2 e3x . The particular integral is obtained   by putting 1 1 1 3x y= e3x , e ; − (D − 2) (D − 1) D−2 D−1 according to the method of resolution into partial fractions. Operate with the first symbolic factor, as above, Z Z 1 y1 = e2x e−2x e3x dx − ex e−x e3x dx = e3x , 2 The complete solution is, therefore, y = C1 ex + C2 e2x + 21 e3x . 2 (2) Solve (D − 2) y = ex . Ansr. y = C1 e2x + C2 xe2x + ex .

Case 3 (Special). When R is a rational function of x, say xn . This case is comparatively rare. The procedure is to expand. f (D)−1 in ascending powers of D as far as the highest power of x in R. 311

There are other (sometimes easier) methods of finding particular integrals. See the supplement starting at the bottom of page S-23 for details. KH

EXAMPLES.–(1) Solve d2 y/dx2 − 4 dy/dx + 4y = x2 . The complementary function is y = e2x (A + Bx); the particular integral is:  
D 1 D 1 −2 2x2 + 4x + 3 . (2 − D) x2 = 1 + 2 + 3 2 x2 = 4 2 2 8 (2) If d2 y/dx2 − y = 2 + 5x, y = C1 ex + C2 e−x − 5x − 2.

Case 4 (Special). When R contains an exponential factor, so that R = eax X,

where X may or may not be a function of x and a, has some constant value. i. When X is a, function of x. Since Dn eax = an eax , where n is any positive integer (page 38), we have (page 95) D (eax X) = eax DX + aeax X = eax (D + a) X, and generally, as in Leibnitz’ theorem (page 49), Dn eax X = eax (D + a)n X; Dn eax eax eax ax ∴ X. n = e X; and nX = Dn (D + a) (D + a)

(5)

R The operation D−1 eax X is performed (when X is any function of x) by transplanting −1 eax from the right- to the left-hand side of the operator f (D) and replacing D by (D + a). This will, perhaps, be better understood from the following examples: EXAMPLES.–(1) Solve d2 y/dx2 − 2 dy/dx + y = x2 e3x . The complete solution by −1 page 308, is (C1 + xC2 ) ex + (D − 2D + 1) x2 e3x . From (5), 1 1 x2 e3x = x2 e3x . D2 − 2D + 1 (D − 1) (D − 1) By rule: e3x may be transferred from the right to the left side of the operator provided we replace D by D + 3.

We get

1

2 2 x . (D + 2)   1 2 1 3 3x , e x − x+ 4 2 8

e3x

as the value of the particular integral. −1 (2) Evaluate (D − 1) ex log x. Ansr. xex log x/e.

ii. When X a constant. If X is constant, the operation (5) reduces to 1 ax 1 ax e = e . f (D) f (a)

(6)

The operation f (D) eax X is performed by replacing D + a by a.
EXAMPLES.–(1) Find the particular integral in D2 − 3D + 2 y = e3x . Obviously, 1 1 1 e3x = 2 e3x = e3x . D2 − 3D + 2 3 −3·3+2 2

(2) Show that 41 ex , is a particular integral in d2 y/dx2 + 2 dy/dx + 1 = ex .

312

See page S-25 of the supplement for better notation of (5) KH

An anomalous case arises when a, is a root of f (D) = 0. By this method, we should get for the particular integral of dy/dx − y = ex . ex 1 ex = = ∞ ex . D−1 1−1 The difficulty is evaded by using the method (5) instead of (6). Thus, 1 1 ex = ex · 1 = x ex D−1 D The complete solution is, therefore, y = Cex + x ex . Another mode of treatment is the following: Since a is a root of f (D) = 0, by hypothesis, D − a is a factor of f (D) (page 386). Hence, ∴

1 eax f (D)

f (D) = (D − a) f 0 (D) ; 1 1 1 1 x eax (7) = · 0 eax = · 0 eax = 0 . (D − a) f (D) (D − a) f (a) f (a)

If the root a occurs r times in f (D) = 0, then D − a enters r times into f (D). Therefore, 1 1 1 1 xr eax 1 ax ax eax = · e = · e = . (8) f (D) (D − a)r f 0 (D) (D − a)r f 0 (a) rf 0 (a) 3

EXAMPLES. – Find the particular integrals in, (1) (D + 1) y = e−x . Ansr. 16 x3 e−x . Hint. Replace D
by D − 1. e−x D−3 , etc. See page
312. −1 (2) D3 − 1 y = x ex . Ansr. ex 61 x2 − 31 x . Hint. First get ex (D − 1) x, then x e (1 + D + . . . ) x, etc.

Case 5 (Special). When R contains sine or cosine factors. By the successive differentiation of sin (nx + a),
n
n D 2 sin (nx + a) = −n2 sin (nx + a) . ∗ where n and a are constants.

∴ f D 2 sin (nx + a) = f −n2 sin(nx + a). 1 1 sin (nx + a) = sin (nx + a) . 2 f (D ) f (−n2 ) It can be shown in the same way that, 1 1 cos (nx + a) = cos (nx + a) . 2 f (D ) f (−n2 ) EXAMPLES.–(1) Find the particular integral of

Here,

d3 y/dx3 + d2 y/dx2 + dy/dx + y = sin 2x. 1 1 R = 3 sin 2x = sin 2x. f (D) D + D2 + D + 1 (D2 + 1) + D (D2 + 1)

* The proof resembles a well-known result in trigonometry, § 19:– D (sin nx) = d (sin nx) /dx = n cos nx;

D2 (sin nx) = d2 (sin nx) /dx2 = −n2 cos nx, etc.

313

(9)

(10)

←−


−1 Substitute for D2 = −22 as in (9).
We thus get − 31 (D + 1) sin 2x. Multiply by 2 2 D − 1 and again substitute D = −2 in the result. Thus 1 15

(D − 1) sin 2x, or

1 15

(2 cos 2x − sin 2x)

is the desired result.
(2) Solve d2 y/dx2 − k 2 y = cos mx. Ansr. C1 ekx + C2 e−kx − (cos mx) / m2 + k 2 . (3) If α and β are the roots of the auxillary equation derived from d2 y/dt2 + m dy/dt + n2 y = a sin nt,

(Helmholtz’s equation for the vibrations of a tuning-fork) show that, is the complete solution.

C1 eαt + C2 eβt − (a cos nt) /mn.

An anomalous case arises when D 2 in D 2 + n2 is equal to −n2 . For instance, the complementary function of d2 y/dx2+n2 y = cos nx, is C1 cos nx+ C2 sin nx, the particular integral is (D 2 + n2 ) cos nx. If the attempt is made to evaluate this, by substituting D 2 = −n2 , we get (cos nx) / (−n2 + n2 ) = ∞ cos nx. We were confronted with a similar difficulty on page 243. The treatment is practically the same. We take the limit of (D 2 + n2 ) cos nx, when n becomes n + h and h converges towards zero. 1 1 cos (n + h) x. cos nx; or 2 −n2 + n2 − (n + h) + n2 1 1 cos (nx + hx) ; ∴ lim cos (n + h) x = lim h→0 − (n + h)2 + n2 h→0 −2nh − h2 1 (cos nx · cos hx − sin nx · sin hx) ; = lim − h→0 2nh + h2     1 h2 x2 = lim − cos nx 1 − + . . . − sin nx (hx − . . . ) ; h→0 2nh + h2 2!  1  cos nx = lim − − x sin nx + powers of h . h→0 2n + h h

But cos nx is contained in the complementary function and hence, when h = 0, we obtain, x sin nx + (a term in the complementary function) . 2n This latter may be disregarded when the particular integral alone is under consideration. The complete solution is, therefore, y = C1 cos nx + C2 sin nx + (x sin nx) /2n. EXAMPLES.–(1) Show that − 21 x cos x, is the particular integral of
D2 + 1 y = sin x
−1 (2) Evaluate D2 + 4 cos 2x. Ansr. 14 x sin 2x.
−1 (3) Evaluate D2 + 4 sin 2x. Ansr. 14 x cos 2x. (4) Solve d3 y/dx3 − y = x sin x. The particular integral consists of two parts, 1 {(x − 3) cos x − x sin x}. The complementary function is 2 √
√
C1 ex + C2 e−x/2 sin 12 3x + C3 ex/2 cos 21 3x .

But see next case.

Case 6. When R contains some power of x as a factor. Say, R = nX, 314

where X is any function of x. The successive differentiation of two products gives, § 20, D n xX = xD n X + nD n−1 X. ∴ f (D) xX = xf (D) X + f 0 (D) X.

Substitute Y = f (D) X, where Y is any function of x. Operate with f (D)−1 , we get   1 1 1 0 xY = x − · f (D) Y. (11) f (D) f (D) f (D)

See page S-31 of the supplement for the Laplace method, which is a single approach to solving all 6 of the cases in this section and more. § 20 is included following the supplement. KH

EXAMPLES.–(1) Find the particular integral in d3 y/dx3 − y = xe2x . From (11), the particular integral is     1 1 1 1 2x 2x = x− 3 · 3D2 e = x − · 3 · 4 e , etc. D −1 D3 − 1 7 7
(2) Show in this
way, that the particular integral of D4 − 1 y = x sin x, is 81 x2 cos x − 3x sin x . (3) Solve d2 y/dx2 − y = x ex sin x. 1 x Ansr. y = C1 ex + C2 e−x − 25 e {(10x + 2) cos x + (5x − 14) sin x}.
2 2 (4) Integrate d y/dx −y = x2 cos x. Ansr. y = C1 ex +C2 e−x +x sin x+ 12 cos x 1 − x2 .

§ 131. The Linear Equation with Variable Coefficients. Case 1. The homogeneous linear differential equation. The general type of this equation is: n−1 dn y y n−1 d + a x + · · · + an y = X, (1) 1 n n−1 dx dx where X is a function of x; a1 , a2 , . . . , an are constants. This equation can be transformed. into one with constant coefficients by the substitution of

xn

x = ez ; or z = log x. we then have, dy/dx = ez and therefore, x dy/dx = dy/dz. (2) Just as we have found it very convenient to employ the symbol, “D” to d ,” so we shall find it even more convenient to denote the operation, “ dx d ,” by the symbol, “ϑ.” “ϑ” is treated in exdenote the operation, “x dx actly the same manner as we have treated “D”* in § 128 and subsequently. * A little care is required in using this new notation. The operations of differentiation and multiplication by a variable are not commutative. The operation x2 D2 is not the same as ϑ2 , or as xD · xD. But we must write, xDy = ϑy;

2

x D2 y = ϑ (ϑ − 1) y;

x3 D3 y = ϑ (ϑ − 1) (ϑ − 2) y; ... ... n n x D y = ϑ (ϑ − 1) (ϑ − 2) . . . (ϑ − n + 1) y.

315

d = EXAMPLES.–(1) ϑ = xD = x dx m m (2) Show that ϑx = mx

d dz .

i. The complementary function. From the first of equations (2), we have § 12, 9,   dy dy dz 1 dy d2 y 1 d2 y dy ; etc. = · = · ; = − dx dz dx x dx dx x dx2 dz Substitute these values in (1). The equation reduces to one with constant coefficients which may be treated by the methods described in the preceding sections. EXAMPLES.–(1) Solve x3 · d3 y/dx3 + 3x · dy/dx − 3y = x2 + x.

∴ ϑ (ϑ − 1) (ϑ − 2) y + 3ϑy − 3y = e2z + ez .
∴ (ϑ − 1) ϑ2 + 3 y = e2z + ez .  √   √  1 1 ∴ y = C1 ez + C2 cos z 3 + C3 sin z 3 + e2z + zez . 7 4  √  1 √ 1 ∴ y = C1 x + C2 cos 3 log x + C3 sin 3 log x + x + x log x. 7 4
(2) Solve x2 · d2 y/dx2 + x · dy/dx + q 2 y = 0; i.e., ϑ2 + q 2 y = 0. Ansr. y = C1 sin (q log x) + C2 cos (q log x).

The linear equation with variable coefficients bears the same relation to x, that the equation with constant coefficients does to emx . Hence if xm be substituted for y, the factor xm will divide out from the result and an equation in m will remain. The n roots of this latter equation will determine the complementary function. EXAMPLES.–(1) Solve x · d2 y/dx2 + 2x · dy/dx − 2y = 0. Put y = xm . We get m (m − 1) + 2 (m − 1) = 0; or

(m + 2) (m − 1) = 0.

Hence from our preceding results, we can write down the complementary function at sight, y = C1 x + C2 x−2 . (2) Solve x2 · d2 y/dx2 + 4x · dy/dx + 2y = 0. Ansr. y = C1 /x + C2 /x2 . (3) Find the complementary function in {ϑ (ϑ − 1) − 3ϑ + 4} y = x3 . Ansr. y = (C1 + C2 log x) x2 . (4) Integrate {ϑ (ϑ − 1) − 2} y = 0. Ansr. y = C1 x2 + C2 /x.

ii. The particular integral. We may use the operator ϑ, to obtain the particular internal of linear equations with variable coefficients in the same way that D was used to determine the particular integral of equations with constant coefficients. The symbolic form of the particular integral is, R y= . f (ϑ) The operator f (ϑ)−1 may be resolved into partial fractions or into factors as in the case of D. 316

EXAMPLES.–(1) Show that y = C1 x4 + C2 /x + 51 x4 log x a complete solution of x · d2 y/dx2 − 2x · dy/dx − 4y = x4 . 1 x3 . Using the ordinary method just described we get (2) Find the value of ϑ2 (ϑ−3) 2

the indeterminate form write

3

1 9

x · ϑ−3 . In this case we must adopt the method of page 312 and Z 1 3 1 dx 1 3 1 x ·1 = x = x3 log x. 9 ϑ−1 9 x 9

(3) Solve x3 · d2 y/dx2 + 7x · dy/dx + 5y = x5 . Write this {ϑ (ϑ − 1) + 7ϑ} y = x5 .


−1 5 The particular integral is ϑ2 + 6ϑ + 5 x , or x5 /60. complementary function is −1 −5 C1 x + C2 x . (4) Solve x2 · d2 y/dx2 + 4x · dy/dx + 2y = ex . Ansr. y = C1 /x + C2 /x2 + ex /x2 .

(5) Solve x3 · d3 y/dx3 + 2x2 d2 y/dx2 − x · dy/dx + y = x + x3 . 2

Ansr. y = C1 /x + C2 x + C3 x log x + 14 x (log x) + x3 /16.

(6) Solve x3 · d3 y/dx3 + 2x2 · d2 y/dx2 + 2y = 10x + 10/x. Ansr. y = C1 x cos (log x) + C2 x sin (log x) + 5x + C3 /x + (2 log x) /x. (7) Find the particular integral of the third example in the last set. Ansr. x3 . (8) Equate example (2), of the preceding set, to 1/x instead of to zero, and show that the particular integral is then (log x) /x.

Case 2. Legendre’s Equation. Type: (a + bx)n

n−1 dn y y n−1 d + A (a + bx) + · · · + An y = R, 1 n n−1 dx dx

(3)

where A1 , A2 , . . . An are constants, R is any function of x. This sort of equation is easily transformed into the homogeneous equation and, therefore, into the linear equation with constant coefficients. To make the former transformation, substitute z = ax + b, for the latter, e = a + bx. EXAMPLES.–(1) Solve 2

(a + bx) · d2 y/dx2 + b (a + bx) · dy/dx + c2 y = 0. Ansr. y = C1 sin {(c/b) log (a + bx)} + C2 cos {(c/b) log (a + bx)}. 2

(2) Solve (x + a) · d2 y/dx2 − 4 (x + a) · dy/dx + 6y = x. 2

2

Ansr. y = C1 (x + a) + C2 (x + a) +

1 6

(3x + 2a).

§ 132. The Exact Linear Differential Equation. A very simple relation exists between the coefficients of an exact differential equation which may be used to test whether the equation is exact or not. Take the equation, d2 y dy d3 y + X3 y = R. (1) X0 3 + X1 2 + X2 dx dx dx 317

Here the “latter” appears to be (3) and the first example but the “former” not to have an example shown. Other than choice of symbols, the two substitutions are equivalent. KH

where X1 , X2 , . . . , R are functions of x. Let their successive differential coefficients be indicated by dashes, thus X 0 , X 00 , . . . Since X0 ·d3y/dx3 has been obtained by the differentiation of X0 ·d2y/dx2 , this latter is necessarily the first term of the integral of (1). But,   2 d2 y d3 y d 0d y X0 2 = X0 3 + X0 2 . dx dx dx dx Subtract the right-hand side of this equation from (1), (X1 − X00 )

dy d2 y + X2 + X3 y = R. 2 dx dx

(2)

Again, the first term of this expression is a derivative of (X1 − X00 ) dy/dx. This, therefore, is the second term of the integral of (1). Hence, by differentiation and subtraction, as before, (X2 − X10 + X000 )

dy + X3 y = R. dx

(3)

This equation may be deduced by the differentiation of (X2 − X10 + X000 ) y, provided the first differential coefficient of (X2 − X10 + X000 ) with respect to x, is equal to X3 that is to say, X20 − X100 + X0000 = X3 ; or X3 − X20 + X100 − X0000 = 0.

(4)

But if this is really the origin of (3), the original equation (1) has been reduced to a lower order, namely, X0

R dy d2 y + (X1 − X00 ) + (X2 − X10 + X000 ) y = R dx + C. 2 dx dx

(5)

This equation is called. the first integral of (1), because the order of the original equation has been lowered unity, by a process of integration. Condition (4) is a test of the exactness of a differential equation. If the first integral is an exact equation, we can reduce it, in the same way, to another first integral of (1). The process of reduction may be repeated until an inexact equation appears, or until y itself is obtained. Hence, an exact equation of the nth order has n independent first integrals. EXAMPLES.–(1) Is the equation x5 · d3 y/dx3 + 15x4 · d2 y/dx2 + 60x3 · dy/dx + 60x2 y = ex exact? From (4), X3 = 60x2 ; X20 = 180x2 ; X100 = 180x2 ; X0000 = 60x2 . Therefore, X3 − X20 + X100 − X0000 = 0 and the equation is exact.
(2) Solve the equation, x · d3 y/dx3 + x2 − 3 d2 y/dx2 + 4x · dy/dx + 2y = 0, as far as possible, by successive reduction. The process can be employed twice, the residue is a linear equation of the first order, not exact. (3) Solve the equation given in example (1). Ansr. x5 y = ex + C1 x2 + C2 x + C3 .

318

See supplement page S-39 for clarification of how the first integral arises. KH

There is another quick practical test for exact differential equations (Forsyth) which is not so general as the preceding. When the terms in X are either in the form of axm , or of the sum of expressions of this type, xm dn y/dxn is a perfect differential coefficient, if m < n. This coefficient can then be integrated whatever be the value of y. If m = n or m > n, the integration cannot be performed by the method for exact equations. To apply the test, remove all the terms in which m is less than n, if the remainder is a perfect differential coefficient, the equation is exact and the integration may be performed. EXAMPLES.–(1) Apply the test to x3 · d4 y/dx4 + x2 · d3 y/dx3 + x · dy/dx + y = 0. x · dy/dx + y remains. This has evidently been formed by the operation D (xy), hence the equation is a perfect differential. (2) Apply the test to
x3 D4 + x2 D3 + x2 D + 2x y = sin x.
x2 · dy/dx + 2xy remains. This is a perfect differential, formed from D x2 y . The equation is exact.

If two independent first integrals are known the equation is sometimes easily solved. The elimination of dy/dx between two first integrals will give the complete solution. § 133. The Integration of Equations with Missing Terms. Differential equations with missing letters are common. First, the independent variable is absent. Type: d2 y/dx2 = qy; or d2 y/dx2 = q f (y) .

(1)

This equation is, in general, neither linear nor exact. Case 1. When f (y), in (1), is negative, so that d2 y + q 2 x = 0, dx2

(2)

where the academic x and y have given place to t and x respectively, in order to give the equation the familiar form of the equation of the motion of a particle under the influence of a central attracting force. Multiply both sides of the equation by 2dx/dt, and integrate with respect to x,  2 dx dx d2 x 2 dx = −q 2 x2 + C. 2 · 2 = −2q x ; or dt dt dt dt Separate the variables and integrate again, √

dx = ±q dt; α2 − x2

cos−1 319

x = ± (qt + ) , α

e.g., if between the 1st integral and the 1st integral of the 1st integral of the equation... KH

where  is the integration constant and C = q 2 α2 . The solution involves two arbitrary constants α and , which respectively denote the amplitude and epoch of a simple harmonic motion, whose period of oscillation is 2π/q. Put C1 = −α cos , and C2 = −α sin . x = C1 cos qt + C2 sin qt. [Case 2.] When f (y) in (1) is positive, the solution assumes the form, x = C1 eqt + C2 e−qt ; or x = A cosh qt + B sinh qt, as on page 309. All these results are important in connection with alternating currents and other forms of harmonic motion. Another way of treating equations of type (9), occurs with an equation like  2 d2 y dy y 2+ − 2y 2 = 0, (3) dx dx which has the form of the standard. equation for the small oscillations of a pendulum in air. Under this condition, the resistance of the air is negligible. Let p = dy/dx, ∴ d2 y/dx2 = dp/dx = p · dp/dy. Substitute these results in the given equation, multiply through with 2/y. py

dp 2 dp + p2 = 2y 2; or 2p + p2 = 4y. dy dy y

Multiply by y 2 and d (y 2p2 ) = 4y 3; or p2 y 2 = y 4 + C 4 , dy where C 4 is an arbitrary constant. The rest is obvious. EXAMPLES.–(1) The solution of equation (3) is sometimes written in the form y 2 = C12 sinh (2x + C2 ) . Verify this. (2) Solve d2 x/dt2 + µx + ν = 0. Put x = x1 + ν/µ and afterwards omit the suffix. √ √ Ansr. x = ν/µ + C1 cos t µ + C2 sin t µ. (3) If the term µx in the preceding example had been of opposite sign, show that √ √ the solution would have been, x = ν/µ + C1 cosh t µ + C2 sinh t µ, where µ is negative. (4) Solve d2 y/dx2 − a (dy/dx)2 = 0. Ansr. C1 x + C2 = e−ay . 2 (5) Solve 1 + (dy/dx) = y d2 y/dx2 . Ansr. y = a cosh (x/a + b). (6) Fourier’s equation for the propagation of heat in a cylindrical bar, is d2 V /dx2 − β 2 V = 0. Hence show that V = C1 eβx + C2 e−βx .

Second, the dependent variable is absent. Type: d2 y/dx2 = x; or d2 y/dx2 = f (x) . 320

(4)

If these equations are exact, they may be solved by successive integration. If the equation has the form d2 y/dx2 + dy/dx + x = 0. d2 v 1 dv P + · =− , 2 dr r dr lµ

Say,

a familiar equation in hydrodynamics, it is usually solved by substituting y = dp/dx, ∴ dp/dx = d2 p/dx2 . The resulting equation is of the first order, integrable in the usual way. EXAMPLES.–(1) The above equation represents the motion of a fluid in a cylindrical tube of radius r and length l. The motion is supposed to be parallel to the axis of the tube and the length of the tube very great in comparison with its radius r. P denotes the difference of the pressure at the two ends of the tube. If the liquid wets the walls of the tube, the velocity is a maximum at the axis of the tube and gradually diminishes to zero at the walls. This means that the velocity is a function of the distance (r1 ) of the fluid from the axis of the tube. Solve the equation, remembering that µ is a constant depending on the nature of the fluid. Substitute p = dv/dr, ∴ dp/dr + p/r = −P/lµ; P P 2 dp + p = − r, is pr = − r + C1 ; r dr lµ 2lµ P C1 P 2 dv =− r+ ; v=− r + C1 log r + C2 . ∴ dr 2lµ r 4lµ

(5)

To evaluate C1 in (5), note that at the axis of the tube r = 0. This means that if C1 , is a finite or an infinite magnitude the velocity will be infinite. This is obviously impossible, therefore, C1 , must be zero. To evaluate C2 , note that when r = r1 , v vanishes and, therefore, we get the final solution of the given equation in the form,
v = 14 P r12 − r2 /lµ, which represents the velocity of the fluid at a distance r1 , from the axis. q 2 (2) Solve ad2 y/dx2 = 1 + (dy/dx) . Make the necessary substitutions and integrate.   p p a · dp/ (1 + p2 ); becomes x/a log p + p2 + 1 + C; or, in the exponential form, p C1 ex/a − p = (1 + p2 );

and dy/dx =

1 C1 ex/a − e−x/a /2C1 ; 2

by squaring. On integration

y=

1 1 aC1 ex/a + ae−x/a /C1 + C2 . 2 2

(3) Some expressions can be reduced to the standard form by an obvious transformation. Thus, d5 y/dx5 − d3 y/dx3 = x. Substitute p for d3 y/dx3 and differentiate p = d3 y/dx3 twice. Thus, d2 p/dx2 − p = x, whence y can be obtained by successive integration as indicated above. (4) Solve d2 V /dr2 + 2dV /r · dr = 0. This equation occurs in the theory of

321

potential. Put dV /dr for the independent variable and divide through. On integration log (dV /dr) + 2 log r = log C1 . where log C1 is an arbitrary constant. Integrate again dV /dr = C1 /r,

becomes V = C + 2 − C1 /r.

(5) If x · d2 y/dx2 = 1, show that y = x log x + C1 x + C2 .

§ 134. Equations of Motion, chiefly Oscillatory Motion. By Newtons second law, if a certain mass (m) of matter is subject to a constant force (F0 ) for a certain time; we have, in rational units, F0 = (Mass) × (Acceleration of the particle). If the motion of the particle is subject to friction, we must regard the friction as a force tending to oppose the motion generated by the impressed force. But friction is proportional to the velocity (v) of the motion of the particle, and equal to the product of the velocity and a constant called the coefficient of friction, written, say, µ. Let F1 denote the total force acting on the particle in the direction of its motion, F1 = F0 − µv = m d2 s/dt2 .

(1)

If there is no friction, we have, for unit mass, F0 = d2 s/dt2 .

(2)

The motion of a pendulum in a medium which offers no resistance to its motion, is that of a material particle under the influence of a central force (F ) attracting with an intensity which is proportional to the distance of the particle away from the centre of attraction. That is (Fig. 7), F = −q 2 s.

(3)

where q 2 is to be regarded as a positive constant which tends to restore the particle to a position of equilibrium – the so-called coefficient of restitution. It is written in the form of a power to avoid a root sign later on. The negative sign shows that the attracting force (F ) tends to diminish the distance (s) of the particle away from the centre of attraction. If s = 1, q 2 represents the magnitude of the attracting force unit distance away. From (2), d2 s = −q 2 s (4) 2 dt This is a typical equation of harmonic motion, as will be shown directly. One solution of (4) is s = C cos (at + ) . (5) 322

This equation is the simple harmonic motion of § 50, C denotes the amplitude of the vibration. If  = 0, we have the simpler equation, s = C cos qt.

(6)

When the particle is at its greatest distance from the central attracting force, qt = π, § 50, page 112. For a complete to and fro motion, 2t = T0 = period of oscillation, hence T0 = 2π/q.

(7)

Equation (4) represents the small oscillations of a pendulum; also the undamped* oscillations of the magnetic needle of a galvanometer. In the sine galvanometer, the restitutional force tending to restore the needle to a position of equilibrium, is proportional to the sine of the angle of deflection of the needle. If J denotes the moment of inertia of the magnetic needle and G the directive force exerted by the current on the magnet, the equation of motion of the magnet, when there is no retarding force, is d2 φ J 2 = −G sin φ. (8) dt For small angles of displacement, φ and sin θ are approximately equal. Hence, d2 φ G = − φ. 2 dt J From (4), q =

p J/G, and therefore, from (9), p T0 = 2π J/G,

(9)

(10)

a well-known relation showing that the period of oscillation of a magnet in the magnetic field, when there is no damping action exerted on the magnet, is proportional to the square root of the moment of inertia of the magnetic needle, and inversely proportional to the square root of the directive force exerted by the current on the magnet. See page 524. In a similar manner, it can be shown that the period p of the small oscillations of a pendulum suspended freely by a string of length l, is 2π l/g, where g denotes the acceleration of gravity.

Equation (4) takes no account of the resistance to which a particle is subjected as it moves through such resisting media as * When an electric current passes through a galvanometer, the needle is deflected and begins to oscillate about a new position of equilibrium. In order to make the needle come to rest quickly, so that the observations may be made quickly, some resistance is opposed to the free oscillations of the needle either by attaching mica or aluminum vanes to the needle so as to increase the resistance of the air, or by bringing a mass of copper close to the oscillating needle. The currents induced in the copper by the motion of the magnetic needle, react on the moving needle, according to Lenz’s law, so as to retard its motion. Such a galvanometer is said to be damped. When the damping is sufficiently great to prevent the needle oscillating at all, the galvanometer is said to be “dead beat” and the motion of the needle is aperiodic. In ballistic galvanometers, there is very much damping.

323

§ 50 reviews the properties of sinusoids as they can be used to describe periodic sinusoidal motion.

air, water, etc. This resistance is proportional to the velocity, and has a negative value. To allow for this, equation (4) must have an additional negative term. We thus get d2 s ds = −µ − q 2 s, dt2 dt where µ, is the coefficient of friction. For greater convenience, we may write this 2f , d2 s ds + 2f + q 2 s = 0. (11) 2 dt dt Before proceeding further, it will perhaps make things plainer to put the meaning of this differential equation into words. The manipulation of the equations so far introduced, involves little more than an application of common algebraic principles. Dexterity in solving comes by practice. Of even greater importance than quick manipulation is the ability to form a clear concept of the physical process symbolised by the differential equation. Some of the most important laws of Nature appear in the guise of an “unassuming differential equation”. The reader should spare no pains to acquire familiarity with the art. The late Professor Tait has said that “a mathematical formula, however brief and elegant, is merely a step towards knowledge, and an all but useless one until we can thoroughly read its meaning.” In equation (11), the term d2 s/dt2 denotes the relative change of the velocity of the motion of the particle in unit time, § 7; 2f · ds/dt shows that this motion is opposed by a force which tends to restore the body to a position of rest, the greater the velocity of the motion, the greater the retardation; q 2 s represents another force tending to bring the moving body to rest, this force also increases directly as the distance of the body from the position of rest. To investigate this motion further, we cannot do better than follow Professor Perry’s graphic method. The first thing is to solve (11) for s. This is done by the method of § 130. Put s = emt and solve the auxillary quadratic equation. We thus obtain p m = f ± f 2 − q2. (12) And finally,

s = e−(α+β)t ; or rather s = C1 e−αt + C2 e−βt , p p where α = −f + f 2 + q 2 and β = −f − f 2 + q 2 . The solution of (11) thus depends on the relative magnitudes of f and q. Suppose that we know enough about the moving system to be able to determine the integration constants. When t = 0, let v = v0 and s = 0.

324

Case i. The roots of the auxillary equation are real and unequal. The condition for real roots −α and −β, in (12), is that f is greater than q (page 388). In this case, s = C1 e−αt + C2 e−βt ,

(13)

solves equation (11). To find what this means, let us suppose that f = 3, q = 2, t = 0, s = 0, v = v0 . From (12), therefore, √ m = −3 ± 9 − 4 = −3 ± 2.24 = −0.76 and − 5.24.

Substitute these values in (18) and differentiate for the velocity v or ds/dt. Thus, s = C1 e−5.24t + C2 e−0.76t ; ds/dt = −5.24C1e−5.24t − 0.76C2 e−0.76t . ∴ −5.24C1 − 0.76C2 = 1. From (18), when t = 0, s = 0 and C1 + C2 = 0, or −C1 = +C2 = 61 .
∴ s = 16 e−0.76t − e−5.24t .

(14)

Assign particular values to t, and plot the corresponding values of s by means of Tables XXI and XXII. Curve No. 1 (Fig. 114) was obtained by plotting corresponding values of s and t obtained in this way.

Figure 114: (after Perry). Case ii. The roots of the auxillary equation are real and equal. The condition for real and equal roots is that f = q. ∴ s = (C1 + C2 t) e−f t .

(15)

As before, let f = 2, q = 2, t = 0, s = 0, v0 = 1. The roots of the auxillary are −2 and −2. Hence s = (C1 + C2 t)e−2t ; and ds/dt = C2 e−2t − 2 (C1 + C2 t) e−2t . ∴ C2 − 2C1 = 1, C1 = 0

and C2 = 1; or s = te−2t .

(16)

Plot (16) in the usual manner. Curve 2 (Fig. 114) was so obtained. Case iii. The roots of the auxillary equation are real, equal and of opposite sign. For equal roots of opposite sign, say ±q, we must have f = 0. Then s = C1 sin qt + C2 cos qt.

Let t = 0, s = 0, v0 = 1, q = 2, f = 0. Differentiate (17), ds/dt = qC1 cos qt − qC2 sin qt.

325

(17)

It would appear that the text is referring to when the roots are purely imaginary (multiples of ι) and opposite sign. KH

Hence 1 = 2C1 × 1 − 2 × C2 × 0, or C1 = 21 ; ∴ C2 = 0. Hence the equation, s=

1 2

sin 2t.

(18)

A graph from this equation is shown in curve 4 (Fig. 114). Case iv. The roots of the auxillary equation are imaginary. For imaginary roots, p f + f 2 − q 2 , or, say a ± bι, it is necessary that f < q (page 388). In this case, s = e−at (C1 sin bt + C2 cos bt) .

(19)

Let the coefficient of friction, f = 1, q√= 2, t = 0, s = 0, v0 = 1.√ The roots of the √ auxillary are m = −1 ± 1 − 4 = −1 ± −3 = −1 + 1.7ι, where ι = −1. Hence a = 1, b = 1.7. Differentiate (19), ds/dt = −ae−at (C1 sin bt + C2 cos bt) + be−at (C1 sin bt − C2 cos bt) . From (19), C2 = 0 and, therefore, C1 = 1/b = −0.57. Therefore, s = 0.57e−t sin 1.7t.

(20)

Curve 3 (Fig. 114) was plotted from equation (20) in the usual way.

There are several interesting feature about the motions represented by these four solutions of (11), shown graphically in Fig. 114. Curves Nos. 3 and. 4 (Cases iv. and iii.) show the conditions under which the equation of motion (11) is periodic or vibratory. The effects of increased friction due to the viscosity of the medium, is shown very markedly by the lessened amplitude and increased period of curve 3. The net result is a damped vibration, which dies away at a rate depending on the resistance of the medium (2f · v) and on the magnitude of the oscillations (q 2 s). Such is the motion of a magnetic or galvanometer needle affected by the viscosity of the air and the electromagnetic action of currents induced in neighbouring masses of metal by virtue of its motion; it also represents the natural oscillations of a pendulum swinging in a medium whose resistance varies as ihe velocity. Curve 4 represents an undamped oscillation, curve 3 a damped oscillation. Curves 1 and 2 (Cases i. and ii.) represent the motion when the retarding forces are so great that the vibration cannot take place. The needle, when removed from its position of equilibrium, returns to its position of rest after the elapse of an infinite time. (What does this statement mean? Compare with page 329.) Raymond calls this an aperiodic motion. To show that the period of oscillation is augmented by damping. From equation (19) we can show that s = e−at A sin bt.

(21)

§ 50. The amplitude of this vibration corresponds to that value of t for which s has a maximum or a minimum value. These values are obtained in the usual way, by equating the first differential coefficient to zero, hence e−at (b cos bt − a sin bt) = 0. 326

(22)

If we now define the angle φ such that bt = φ, or tan φ = a/b, 1 2π

(23)

◦

φ, lying between 0 and (i.e., 90 ), becomes smaller as a increases in value. We have p just seen that the imaginary roots of −f ± f 2 − q 2 are −a + bι, for values of f less than q. Let a2 + b 2 = q 2 . (24) The period of oscillation of an undamped oscillation is, by (7), T0 = 2π/q, of a damped oscillation T = 2π/b.
∴ T 2 /T02 = q 2 /b2 = a2 + b2 /b2 = 1 + a2 /b2 , p ∴ T /T0 = a2 + b2 /b. (25) which expresses the relation between the periods of oscillation of a damped and of an undamped oscillation. The period of vibration is thus augmented on damping. It is easy to show by plotting that tan φ, of (28), is a periodic function such that Hence

tan φ = tan (φ + π) = tan (φ + 2π) = . . . φ, φ + π, φ + 2π, . . .

satisfy the above equation. It also follows that bt1 , bt2 + π, bt3 + 2π, . . . also satisfy the equation, where t1 , t2 , t3 . . . are the successive values of the time. Hence bt2 = bt1 + π, bt3 = bt1 + 2π, . . . ; 1 ∴ t2 = t1 + T, t3 = t1 + T, . . . 2 Substitute these values in (21) and put s1 , s2 , s3 , . . . for the corresponding displacements, ∴ s1 = Ae−at1 sin bt1 ; − s2 = Ae−at2 sin bt2 ; . . . where the negative sign indicates that the displacement is on the negative side. Hence s1 /s2 = e−a(t1 −t2 ) = eaT /2 ; s2 /s3 = s3 /s4 = · · · = eaT /2 . The amplitude thus diminishes in a constant ratio. values of s and t, we get the curve shown in Fig. 115. This ratio is called the damping ratio, by Kohlrausch (“D¨ ampfungsverhaltnis”). It is written k. The natural logarithm of the damping ratio, is Gauss’ logarithmic decrement, written L (the ordinary logarithm of k, is written L). Hence

(26) Plotting these successive

Figure 115: Damped Oscillation and from (25),

λ = log k = aT log e = aT = aπ/b,   1 λ2 λ2 T2 1 + = 1 + ; or T = T + . . . . · 0 T02 π2 2 π2

327

(27) (28)

Hence, if the damping is small, the period of oscillation is augmented by a small quantity of the second order. The following table contains six observations of the amplitudes of a sequence of damped oscillations:

Observed Deflection

k.

λ.

69 48 33.5 23.5 16.5 11.5 8

1.438 1.434 1.426 1.425 1.435 1.438

0.3633 0.3604 0.3548 0.3542 0.3612 0.3633

L.

0.1578 0.1565 0.1541 0.1538 0.1569 0.1578

Meyer, Maxwell, etc., have calculated the viscosity of gases from the rate at which the small oscillations of a vibrating pendulum are damped.

When the motion represented by equation (11) is subject to some periodic impressed force which prevents the oscillations dying away, the resulting motion is said to be a forced vibration. The equation representing such an oscillation is d2 s ds + + q 2 s = f (x) . 2 dt dt

(29)

When f (x) = 0, the equation refers to the natural oscillations of a vibrating electrical or mechanical system. The impressed force is, therefore, mathematically represented by the particular integral of equation (29) (see example (3) below). The subjoined examples’principally refer to systems in harmonic motion. EXAMPLES.–(1) Ohms law for a constant current is E = RC; for a variable current of C amperes flowing in a circuit with a coefficient of self-induction of L henries, with a resistance of R ohms and an electromotive force of E volts, Ohms law is represented by the equation, E = RC = L · dC/dt, (30) where dC/dt evidently denotes the rate of increase of current per second, L is the equivalent of an electromotive force tending to retard the current. (i.) When E is constant, the solution of (30) has been obtained in a preceding set of examples, C = E/R + Be−Rt/L , where B is the constant of integration. To find B, note that when t = 0, C = 0. Hence,   (31) C = E 1 − e−Rt/L R. The second term is the so-called “extra current at make,” an evanescent

328

Modern texts use the symbol, i, for current instead of C. The latter is customarily used for electrical capacitance rather than current. KH

factor due to the starting conditions. The current, therefore, tends to assume the steady condition: C = E/R, when t is very great. (ii.) When C is an harmonic function of the time, say, C = C0 sin qt;

∴ dC/dt = C0 q cos qt,

or, compounding these harmonic motions (§ 50),

Modern engineering texts use the word, “transient” rather than “evanescent.” KH

E = RC0 sin qt + LC0 q cos qt, where  = tan−1p(Lq/R), the so-called lag* of the current behind the electromotive force, the expression R2 + L2 q 2 is the so-called impedance. (iii.) When E is a function of the time, say f (t), Z 1 C = Be−Rt/L = e−Rt/L e−Rt/L f (t) · dt, L The evanescent term e−Rt/L may be omitted when the current has settled down into the steady state. (Why?) (v.) When E is zero, C = Be−Rt/L . Evaluate the integration constant B by putting C = C0 , when t = 0. (2) The relation between the charge (q) and the electromotive force (E) of two plates of a condensor of capacity C connected by a wire of resistance R, is E = R · dq/dt + q/C, provided the self-induction is zero. Solve for q. Show that when Z 1 −t/RC et/RC f (t) · dt + Be−t/RC ; E = f (t) , q = e R E = 0, q = Q0 e−t/RC ; (Q0 is the charge when t = 0). E = constant, q = CE + Be−t/RC ;
E = E0 sin ωt, q = Be−t/RC + CE0 (sin ωt + RCω cos ωt) / 1 + R2 C 2 ω 2 .

(3) The equation of motion of a pendulum subject to a resistance which varies with the velocity and which is acted upon by a force which is a simple harmonic function of the time, is d2 x dx + 2f + q 2 x = cos (qt + ) . dt2 dt Show that the complementary function is x = A cos (qt + ) + B sin (qt + ) . * An alternating (periodic) current is not always in phase (or, “in step”) with the impressed (electromotive) force driving the current along the circuit. If there is selfinduction in the circuit, the current lags behind the electromotive force; if there is a condensor in the circuit, the current in the condenser is greatest when the electromotive force is changing most rapidly from a positive to a negative value, that is to say, the maximum current is in advance of the electromotive force, there is then said to be a lead in the phase of the current.

329

These equations follow from (2) on page 311 or from the alternative methods detailed in § 130. Review that section if you have difficulty arriving at these solutions. KH

To solve the equation, assume that x = A cos (qt + ) + B sin (qt + ) , is a solution. Substitute in the given equation, ∴ −Aq 2 + 2f Bq + n2 A = 1; A= Put

∴ −Aq 2 + 2f Bq + n2 B = 0.

2f q n2 − q 2 ; B= 2 . 2 2 2 − q ) + 4f q (n − q 2 ) + 4f 2 q 2 A = R cos . B = R sin .

(n2

The solution of the given equation is then x = R cos (qt +  − 1 ) , (32) p
2 2 where R = 1/ (n2 − q 2 ) + 4f 2 q 2 ; tan  = 2f q/ n − q . The forced oscillations due to the impressed periodic force, are thus determined by (32). The complementary function gives the natural vibrations superposed upon these. (4) If the friction in the preceding example, is zero, d2 x + n2 x = cos (qt + ) . (33) dt2
A particular integral is x = {f · cos (qt + )} / n2 − q 2 . This fails when n = q. In this case, assume that x = Ct sin (nt + ) is a particular integral. (33) is satisfied. provided C = f /2n. The physical meaning of this is that when the pendulum is acted on by a periodic force “in step” with the oscillations of the pendulum, the amplitude of the forced oscillations will increase proportionally with the time, until, when the amplitude exceeds a certain limit, equation (33) no longer represents the motion of the pendulum. (5) When an electric current, passing through an electrolytic cell, has assumed the steady state, show that the ionic velocity is proportional to the impressed force (electromotive force). By Newtons law, for a moving body, (Impressed force) = (Mass) × (Acceleration) . Friction is to be regarded as a retarding force acting in an opposite direction to the impressed force; this frictional force is proportional to the velocity of the body. ∴ (Impressed force less friction ) = (Mass) × (Acceleration) . Express these facts in symbolic language. See (1) above. Integrate the result and evaluate the constant for v = 0, when t = 0.   (34) ∴ µv = F 1 − e−µt/m .

For ionic motion, m is very small, µ, is very great. When t is great, show that the exponential term vanishes, and F ∝ v.

§ 135. The Velocity of Simultaneous and Dependent Chemical Reactions. While investigating the rate of decomposition of phosphine (§ 88), we had occasion to point out that the action really takes place in two stages:— STAGE I. STAGE II.

P H3 = P + 3H. 4P = P4 ; 2H = H2 . 330

This paragraph demonstrates the method of undetermined coefficients. Further examples of this method can be found in the supplement pages S-24 through S-29.

The former change alone determines the velocity of the whole reaction. The physical meaning of this is that the speed of the reaction which occurs during stage II., is immeasurably faster than the speed of the first. Experiment quite fails to reveal the complex nature of the complete reaction.* Suppose, for example, a substance A forms an intermediate compound B, and this, in turn, forms a final product C. If the speed of the reaction A = B, is one gram per

1 second, 100000

when the speed of the reaction B = C, is one gram per hour, the observed “order” of the complete reaction A = C, will be fixed by that of the slower reaction, B = C, because the methods used for measuring the rates of chemical reactions are not sensitive to changes so rapid as the assumed rate of transformation of A into B. Whenever the “order” of this latter reaction, B = C is alone accessible to measurement. If, therefore, A = C is of the first, second, or nth order, we must understand. that one of the subsidiary reactions (A = B, or B = C) is (1) an immeasurably fast reaction, accompanied by (2) a slower measurable change of the first, second or nth order, according to the particular system under investigation. If, however, the velocities of the two reactions are of the same order of magnitude, the “order” of the complete reaction will not fall under any simple type (§§ 88, 89), and, therefore, some changes will have to be made in the differential equations representing the course of the reaction. Let us study some of the simpler cases. Case i. In a given system, a substance A forms an intermediate substance B, which finally forms a third substance C. Let one gram molecule of the substance A be taken. At the end of a certain time t, the system contains x of A, y of B, z of C. The rate of diminution of x is evidently −

dx = k1 x, dt

(1)

* Professor Walker illustrates this by the following analogy (“Velocity of Graded Reactions,” Proc. Royal Soc. Edin., Dec., 1897): “The time occupied in the transmission of a telegraphic message depends both on the rate of transmission along the conducting wire and on the rate of the messenger who delivers the telegram; but it is obviously this last, slower rate that is of really practical importance in determining the total time of transmission”. . . .

331

where k1 denotes the velocity constant of the transformation of A to B. The rate of formation of C is dz = k2 y, (2) dt where k2 is the velocity constant of the transformation of B to C. Again, the rate at which B accumulates in the system is evidently the difference in the rate of diminution of x and the rate of increase of z, or dy = k1 x − k2 y. dt

(3)

The speed of the chemical reactions, A = B = C, is fully determined by this set of differential equations. When the relations between a set of variables involves a set of equations of this nature, the result is said to be a system of simultaneous differential equations. In a great number of physical problems, the interrelations of the variables are represented in the form of a system of such equations. The simplest class occurs when each of the dependent variables is a function of the independent variable. The simultaneous equations are said to be solved when each variable is expressed in terms of the independent variable, or else when a number of equations between the different variables can be obtained free from differential coefficients. To solve the present set of differential equations, first differentiate (2), dy d2 z − k2 = 0; dt2 dt Add and subtract k1 k2 y, substitute for dy/dt from (3) and for k2 y from (2), we thus obtain dz d2 z + (k1 + k2 ) − k1 k2 (x + y) = 0. dt2 dt But from the conditions of the experiment, x + y + z = 1,

∴ z − 1 = − (x + y) .

Hence, the last equation may be written, d (z − 1) d2 (z − 1) + (k1 + k2 ) + k1 k2 (z − 1) = 0. dt2 dt

(4)

This linear equation of the second order with constant coefficients, is to be solved for z − 1 in the usual manner (§ 130). At sight, therefore, z − 1 = C1 e−k1 t + C2 e−k2 t .

(5)

∴ C1 + C2 = 1.

(6)

But z = 0, when t = 0, Differentiate (5). From (2), dz/dt = 0, when t = 0. Therefore, making the necessary substitutions, −C1 k1 − C2 k2 = 0. (7)

From (6) and (7),

C1 = k1 / (k2 − k1 ) ; C2 = k2 / (k2 − k1 ) .

The final result may therefore be written, z−1=

k1 k2 e−k2 t + e−k1 t . k1 − k2 k2 − k1

332

(8)

Harcourt and Esson have studied the rate of reduction of potassium permanganate by oxalic acid. 2KM nO4 + 3M nSO4 + 2H2 O = K2 SO4 + 2H2 SO4 + 5M nO2 ; M nO2 + H2 SO4 + H2 C2 O4 = M nSO4 + 2H2 O + 2CO2 . By a suitable arrangement of the experimental conditions this reaction may be used to test equations (5) or (8). Let x, y, z, respectively denote the amounts of M n2 O7 , M nO2 , and M nO (in combination) in the system. The above workers found that C1 = 28.5; C2 = 2.7; e−k1 = 0.82; e−k2 = 0.98. The following table places the above suppositions beyond doubt.

t minutes.

0.5 1.0 1.5 2.0 2.5

t minutes.

z − 1. Found.

Calculated.

25.85 21.55 17.9 14.9 12.55

25.9 21.4 17.8 14.9 12.5

z − 1. Found.

Calculated.

10.45 8.95 7.7 6.65 5.7

10.4 9.0 7.8 6.6 5.8

3.0 8.5 4.0 4.5 5.0

Case ii. A solution contains a gram molecules of each of A and C, the substance A gradually changes to B, which, in turn, reacts with C to form another compound D. Let x denote the amount of A which remains untransformed after the elapse of an interval of time t, y the amount of B, and z the amount of C present in the system after the elapse of the same interval of time t. Hence show that −

dx = k1 x; dt

−

dz = k2 yz. dt

(9)

The rate of diminution of B is proportional to the product of its active mass y into the amount of C present in the solution at the time t, but the velocity of increase of y is equal to the velocity of diminution of x, ∴

dy = k1 x − k2 yz. dt

(10)

If x, y, z, could be measured independently, it would be sufficient to solve these equations as in case i., but if x and y are determined together, we must proceed a little differently. Note z = x+y. From the first of equations (9), and (10) by addition and the substitution of dt = −dx/k1 x from (9), and of z − x = y, we get K K 1 dz · + − = 0. 2 z dt z x

(11)

where K has been written in place of k2 /k1 . The solution of this equation has been previously determined (page 298) in the form   1 2 (Kx) + . . . z = 1. (12) Ke−Kx C1 − log x + Kx − 1.22

333

In some of Harcourt and Esson’s experiments, C1 = 4.68; k1 = 0.69; k2 = 0.006864. From the first of equations (9), it is easy to show that x = ae−k1 t. Where does a come from? What does it mean? Hence verify the third column in the following table: t minutes.

z.

2 8 4 5

Found.

Calculated.

51.9 42.4 85.4 29.8

51.6 42.9 85.4 29.7

After the lapse of six minutes, the value of x was found to be negligibly very small. The terms succeeding log x in (12) may, therefore, be omitted without committing any sensible error. Substitute x = ae−k1 t in the remainder, k2 (C1 log a + k1 t) t; or k1

(C10 + t) z =

1 k20

where C10 = C1 /k1 − (log a) /k1 . Harcourt and Esson found that C10 = 0.1, and 1/k2 = 157. Hence, in continuation of the preceding table, these investigators obtained the results shown in the following table. The agreement between the theoretical and experimental numbers is remarkable.

t minutes.

6 7 8 9

t minutes.

z − 1. Found.

Calculated.

25.7 22.2 19.4 17.8

25.7 22.1 19.4 17.8

10 15 20 30

z − 1. Found.

Calculated.

15.5 10.4 7.8 5.5

15.5 10.4 7.8 5.2

The theoretical numbers are based on the assumption that the chemical change consists in the gradual formation of a substance which at the same time slowly disappears by reason of its reaction with a proportional quantity of another substance. This really means that the so-called “initial disturbances” in chemical reactions, are due to the fact that the speed during one stage of the reaction, is faster than during the other. The magnitude of the initial disturbances depends on the relative magnitudes of k1 and k2 . The observed velocity in the steady state depends on the difference between the steady diminution −dx/dt and the steady rise dz/dt. If k2 is infinitely great in comparison with k1 , (8) reduces to
z = a 1 − e−k1 t .

334

which will be immediately recognised as another way of writing the familiar equation k1 =

1 a log . t a−z

So far as practical work is concerned, it is necessary that the solutions of the differential equations shall not be so complex as to preclude the possibility of experimental verification. Case iii. In a given system A combines with R to form B, B combines with R to form C, and C combines with R to form D. In the hydrolysis of triaeetin, C3 H5 · A3 + H · OH = 3A · H + C3 H5 (OH) 3, (Triacetin) (Glycerol) where A has been written for CH3 · COO·, there is every reason to believe that the reaction takes place in three stages: C3 H5 · A3 + H · OH = A · H + C3 H5 · A2 · OH C3 H5 · A2 · OH + H · OH = A · H + C3 H5 · A · (OH)2 C3 H5 · A · (OH)2 + H · OH = A · H + C3 H5 · (OH)3

(diacetin); (monacetin); (glycerol).

These reactions are interdependent. The rate of formation of glycerol is conditioned by the rate of formation of monacetin; the rate of monacetin depends, in turn, upon the rate of formation of diacetin. There are, therefore, three simultaneous reactions of the second order taking place in the system. Let a denote the initial concentration (gram molecules per unit volume) of triacetin, b the concentration of the water; let x, y, z, denote the number of molecules of mono-, di- and triacetin hydrolysed at the end of t minutes. The system then contains a − z molecules of triaoetin, z−y, of diacetin, y−x, of monacetin, and b−(x + y + z) molecules of water. The rate of hydrolysis is therefore completely determined by the equations: dx/dt = k1 (y − x) (b − x − y − z) ; dy/dt = k2 (z − y) (b − x − y − z) ; dz/dt = k3 (a − z) (b − x − y − z) ;

(13) (14) (15)

where k1 , k2 , k3 , represent the velocity coefficients (§ 88) of the respective reactions. Geitel tested the assumption: k1 = k2 = k3 . Hence dividing (15) by (13) and by (14), he obtained dz/dy = (a − z) / (z − y) ; dz/dx = (a − z) / (y − x) .

(16)

From the first of these equations, dy + y

dz z · dz = . a−z a−z

which can be integrated as a linear equation of the first degree. The constant is equated by noting that if a = 1, z = 0, y = 0. The reader might do this as an exercise on § 123. The answer is y = z + (a − z) log (a − z) . (17) Now substitute (17) in the second of equations (16), rearrange terms and integrate as a further exercise on linear equations of the first order. The final result is, x = z + (a − z) log (a − z) −

335

a−z 2 {log (a − z)} . 2

(18)

See worked solution of this equation on page S-42. KH

Geitel then assigned arbitrary numerical values to z (say from 0.1 to 1.0), calculated the corresponding amounts of z and y from (17) and (18) and compared the results with his experimental numbers. For experimental and other details the original memoir must be consulted (vide infra). EXAMPLE. –Calculate equations analogous to (17) and (18) on the supposition that k1 6= k2 6= k3 .

A study of the differential equations representing the mutual conversion of red into yellow, and yellow into red phosphorus, will be found in a paper by Lemoine in the Annales de Chimie et de Physique [4], 27, 289, 1872. There is also a series of interesting papers by Rud. Wegscheider bearing on this subject in Zeit. f. phys. Chem., 30, 593, 1899; ib., 34, 290, 1900; ib., 35, 513, 1900; Monatshefte f¨ ur Chemie, 22, 749, 1901. The preceding discussion is based upon papers by Harcourt and Esson, Phil. Trans., 156, 198, 1866; Geitel, Journ. f¨ ur prakt. Chem. [2], 55, 429, 1897; J. Walker, Proc. Roy. Soc. Adman., 22, 1897. It is somewhat surprising that Harcourt and Esson’s investigation has not received more attention from the point of view of simultaneous and dependent reactions. The indispensable differential equations, simple as they are, might perhaps account for this. But chemists, in reality, have more to do with this type of reaction than any other. The day is surely past when the study of a particular reaction is abandoned simply because it “won’t go” according to the stereotyped velocity equations of § 88. § 136. Simultaneous Differential Equations. By way of practice it will be convenient to study a few more examples of simultaneous equations. For a complete determination of each variable there must be the same number of equations as there are independent variables. Quite an analogous thing occurs with the simultaneous equations of ordinary algebra. I. Simultaneous equations with constant coefficients. The methods used for the solution of these equations are analogous to those employed for similar equations in algebra. The operations here involved are chiefly processes of elimination and substitution, supplemented by differentiation or integration at various stages of the computation. The use of the symbol D often shortens the work. Most of the following examples are from results proved in the regular textbooks on physics. EXAMPLES.–(1) Solve dx/dt + ay = 0, dy/dt + bx = 0. Differentiate the first, multiply the second by a. Subtract and y disappears. Hence writing p p x = C1 emt + C2 e−mt ; or, y = C2 b/a · e−mt − C1 b/a · emt .

336

We might have obtained an equation in y, and substituted it in the second. Thus four constants appear in the result. But one pair of these constants can be expressed in terms of the other two. Two of the constants, therefore, are not arbitrary and independent, while the integration constant is arbitrary and independent. It is always best to avoid an unnecessary multiplication of constants by deducing the other variables from the first without integration. The number of arbitrary constants is always equal to the sum of the highest orders of the set of differential equations under consideration. (2) Solve dx/dt + y = 3x; dy/dt − y = x. Differentiate the first. Subtract each
of the given equations from the result. D2 − 4D + 4 x remains. Solve as usual. x = (C1 + C2 x) e2t . Substitute this value of x in the second of the given equations and y = (C1 − C2 + C3 t) e2t . (3) The equations of rotation of a particle in a rigid plane, are dx/dt = µy; dy/dt = −µx. To solve these, differentiate the first, multiply the second by µ, etc. Finally x = C1 cos µt + C2 sin µt; y = C10 cos µt + C20 sin µt. To find the relation between these constants, substitute these values in the first equation and = −µC1 sin µt + µC2 cos µt = µC10 cos µt + µC20 sin µt, or C10 = C2 and C20 = −C1 . (4) Solve d2 x/dt2 = −n2 x; d2 y/dt2 = −n2 y. x = C1 cos nt + C2 sin nt; y = C10 cos nt + C20 sin nt. Eliminate t so that 2

2

2

(C10 x − C1 y) + (C20 x − C2 y) = (C1 C20 − C2 C10 ) , etc. The result represents the motion of a particle in an elliptic path, subject to a central gravitational force. (5) Solve dx/dt + by + cz = 0; dy/dt + a1 x + c1 z = 0; dz/dt + a2 x + b2 z = 0. Operate on the first with D2 − b2 c1 on the second with b2 c − bD, on the third with bc1 − cD. Add. The terms in y and z disappear. The remaining equation has the integral, x = C1 eαt + C2 eβt + C3 eγt , where α, β, γ, are the roots of z 3 + (a1 b + a2 c + b2 c1 ) z + a1 b2 c + a2 bc1 = 0. The values of y and z are easily obtained from that of x by proper substitutions in the other equations. (6) If two adjacent circuits have currents i1 and i2 then, according to the theory of electromagnetic induction, M

di2 di1 di2 di1 + L2 + R2 i2 = E2 ; M + L1 + R1 i1 = E1 , dt dt dt dt

(see J. J. Thomsons Elements of Electricity and Magnetism, p. 882), where R1 , R2 denote the resistances of the two circuits, L1 , L2 the coefficients of self-induction, E1 , E2 the electromotive forces of the respective circuits and M the coefficient of mutual induction. All the coefficients are supposed constant. First, solve these equations on the assumption that E1 = E2 = 0. Assume that i1 = aemt

and i2 = bemt ,

337

←−

satisfy the given equations. Differentiate each of these variables with respect to t and substitute in the original equation aM m + b (L2 m + R2 ) = 0; bM m + a (L1 m + R1 ) = 0. multiply these equations so that (L1 L2 − M ) m2 + (L1 R2 + R1 L2 ) m + R1 R2 = 0.

For physical reasons, the induction L1 L2 must always be greater than M ; The roots of this quadratic must, therefore, be negative and real (page 388), and i1 = a1 e−m1 t , or a2 e−m2 t ; i2 = b1 e−m1 t , or b2 e−m2t . Hence, from the preceding equation, a1 M m1 + b1 L2 m1 + B1 R2 = 0; or

a1 /b1 = (L2 m1 + R2 ) /M m1 ;

similarly a2 /b2 = M m2 / (L1 M2 + R1 ). Combining the particular solutions for i1 and i2 , we get i1 = a1 e−m1 t + a2 e−m2 t ; i2 = b1 e−m1 t + b2 e−m2 t , the required solutions. Second, if E1 and E2 , have some constant value, i1 = E1 /R1 + a1 e−m1 t + a2 e−m2 t ; i2 = E2 /R2 + b1 e−m1 t + b2 e−m2 t , are the required solutions.

II. Simultaneous equations with variable coefficients. The general type of simultaneous equations of the first order, is  P1 dx + Q1 dy + R1 dz; (1) P2 dx + Q2 dy + R2 dz, . . .

where the coefficients are functions of x, y, z. These equations can often be expressed in the form dy dz dx = = , (2) P Q R which is to be looked upon as a typical set of simultaneous equations of the first order. If one of these equations involves only two differentials, the equation is to be solved in the usual way, and the result used to deduce values for the other variables, as in the first of the subjoined examples. When the members of a set of equations are symmetrical, the solution can often be simplified by taking advantage of a well-known theorem* in algebra (ratio). According to this, * Perhaps it is best to state the proof. Let dx/P = dy/Q = dz/R = k, say; then, dx = P k; dy = Qk; dz = Rk; or, l dx − lP k; m dy − mQk; n dz = nRk. Add these results, l dx + m dy + n dz = k (lP + mQ + nR) . dx dy dz l dx + m dy + n dz =k= = = . ∴ lP + mQ + nR P Q R

338

dx dy dz l dx + m dy + n dz l0 dx + m0 dy + n0 dz = = = = 0 = P Q R lP + mQ + nR l P + m0 Q + n0 R

(3)

where l, m, n, l0 , m0 , n0 ,. . . are sets of multipliers such that hence,

lP + mQ + nR = 0; l0 P + m0 Q + n0 R = 0, etc. l dx + m dy + n dz = 0, etc.

(4) (5)

The same relations between x, y, z, that satisfy (5), satisfy (2). If (4) be an exact differential equation, equal to say du, direct integration gives the integral of the given system, viz., u = a,

(6)

where a, denotes the constant of integration. In the same way, if l dx + m dy + n dz = 0, is an exact differential equation, equal to say dv, then, since dv is also equal to zero, v = b, (7) is a second solution. These two solutions must be independent. EXAMPLES.–(1) Solve dx/y = dy/x = dz/z. The relation between dx and dy contains x and y only, the integral, y 2 − x2 = C1 , follows at once. Use this result to eliminate x from the relation between dy and dz. The result is p p dz/z = dy/ y 2 − C1 ; or, y + y 2 + C1 = C2 z.

These two equations, involving two constants of integration, constitute a complete solution. (2) Solve dx/ (mz − ny) = dy/ (nz − lz) = dz/ (ly − mx). l, m, n and x, y, z form a set of multipliers satisfying the above condition. Hence, l dx + m dy + n dz = 0; x dx + y dy + z dz = 0.

The integrals of these equations are u = lx + my + nz = C1 ; v = x2 + y 2 + z 2 = C2 , which constitute a complete solution.
(3) Solve dx/ x2 − y 2 − z 2 = dy/2xy = dz/2xz. From the two last equations y = C1 z. Substituting x, y, z for l, m, n, each of the given ratios is equal to
(x dx + y dy + z dz) / x2 + y 2 + z 2 . ∴ x2 + y 2 + z 2 = c2 z.

is another solution.

§ 137. Partial Differential Equations. Equations obtained by the differentiation of functions of three or more variables are of two kinds: 1. Those in which there is only one independent variable, such as P dx + Q dy + R dz = S dt which involves four variables – three dependent and one independent. These are called total differential equations. 339

2. Those in which there is only one dependent and two or more independent variables, such as, P

∂z ∂z ∂z +Q +R = 0, ∂x ∂y ∂t

where z is the dependent variable, x, y, t the independent variables. These equations are classed under the name partial differential equations. The former class of equations are rare, the latter very common. We shall confine our attention to partial differential equations. In the study of ordinary differential equations, we have always assumed that the given equation has been obtained by the elimination of constants from the original equation. In solving, we have sought to find this primitive equation.* Partial differential equations, however, may be obtained by the elimination of arbitrary functions of the variables as well as of constants. It can be shown from Euler’s theorem (page 56) that if y z  n u=x f , ,... , x x be a homogeneous function, y z  ∂u ∂u ∂u +y +z + · · · = nxm f , , . . . = nu, x ∂x ∂y ∂z x x

page 56 is included in this document following the supplement.

where the arbitrary function has disappeared.† Again, if
u = f ax3 + by 3 ,

is an arbitrary function of x and y.

∂u ∂u = af 0 ax3 + by 3 ; = bf 0 ax3 + by 3 ; ∂x ∂y

∴ b

∂u ∂u −a = 0. ∂x ∂y

* Physically, the differential equation represents the relation between the dependent and the independent variables corresponding to an infinitely small change in each of the independent variables. The reader will, perhaps, have noticed that the term “independent variable” is an equivocal phrase. (1) If u = f (z), u is a quantity whose magnitude changes when the value of a changes. The two magnitudes u and z are mutually dependent. For convenience, we fix our attention on the effect which a variation in the value of z has upon the magnitude of u. If need be we can reverse this and write z = f (u), so that u now becomes the “independent variable”. (2) If v = f (x, y), x and y are “independent variables” in that x and y are mutually independent of each other. Any variation in the magnitude of the one has no effect on the magnitude of the other. x and y are also “independent variables” with respect to v in the same sense that a has just been supposed the “independent variable” with respect to u. † This is usually proved in the textbooks in the following manner: Let u = xn f (y/x, z/x, . . . ). Put y/x = Y , z/x = Z,. . . ∴ ∂Y /∂x = −y/x2 , ∂Z/∂x = −z/x2 . . . ; ∂Y /∂y = 1/x, ∂Z/∂y = 0, . . .

Let v = f (Y, Z, . . . ), for the sake of brevity, therefore, since u = xn v,

340

←−

∂u EXAMPLES.–(1) If y − bu = f (bx − ay), b ∂u ∂x + b ∂y = 1. (2) If 1/z − 1/x = f (1/y − 1/x), x2 ∂z/∂x + y 2 ∂z/∂y = z 2 . (3) If z = a (x + y), ∂z/∂x − ∂z/∂y = 0.

For this reason an arbitrary function of the variables is added to the result of the integration of a partial differential equation instead of the constant hitherto employed for ordinary differential equations. If the number of arbitrary constants to be eliminated is equal to the number of independent variables, the resulting differential equation is of the first order. The higher orders occur when the number of constants to be eliminated, exceeds that of the independent variables. If u = f (x, y), there will be two differential coefficients of the first order, namely, ∂u/∂x and ∂u/∂y; three of the second order, namely, ∂ 2 u/∂x2 , ∂ 2 u/∂x∂y, ∂ 2 u/∂y 2 . . .

§ 138. What is the Solution of a Partial Differential Equation? Ordinary differential equations have two classes of solutions – the complete integral and the singular solution. Particular solutions are only varieties of complete integral. Three classes of solutions can be obtained from some partial differential equations, still regarding the particular solution as a special case of the complete integral. These are indicated in the following example. The equation of a sphere in three dimensions is, x2 + y 2 + z 2 = r 2 ,

(1)

when the centre of the sphere coincides with the origin of the coordinate planes and r denotes the radius of the sphere. If the centre of the sphere lies somewhere on the xy-plane at a point (a, b), the above equation becomes ∂u = nxn−1 f (Y, Z, . . . ) + xn ∂x



∂v ∂Y ∂v ∂Z · + · + ... ∂Y ∂x ∂Z ∂x



;

by the method for the differentiation of a function of a function, 6 and 9, § 12. Therefore,   ∂v ∂v n−1 n−2 y = nx f (Y, Z, . . . ) − x +z + ... . ∂Y ∂Z ∂v ∂Y ∂v ∂u ∂v ∂Z ∂v ∂u = xn · = xn−1 ; = xn · = xn−1 ;... ∂y ∂dY ∂y ∂Y ∂z ∂Z ∂z ∂Z Now multiply by x, y, z,. . . respectively, and add, x

∂u ∂u ∂u +y +z + · · · = nxn f (Y, Z, . . . ) = nu. ∂x ∂y ∂z

341

(x − a)2 + (y − b)2 + z 2 = r 2 .

(2)

When a and b are arbitrary constants, each or both of which may have any assigned magnitude, equation (2) may represent two infinite systems of spheres of radius r. The centre of any member of either of these two infinite systems (called a double infinite system) must lie somewhere on the xy-plane. Differentiate (2) with respect to x and y. x−a+z

∂z ∂z = 0; y − b + z = 0. ∂x ∂y

Substitute for x − a and y − b in (2). We obtain (  )  2 2 ∂z ∂z z2 + + 1 = r2. ∂x ∂y

(3)

(4)

Equation (2), therefore, is the complete integral of (4). By assigning any particular numerical value to a or b, a particular solution of (4) will be obtained, such is (x − 1)2 + (y − 79)2 + z 2 = r 2 . (5)

If (2) be differentiated with respect to a and b, ∂  (x − a)2 + (y − b)2 + z 2 = r 2 ; ∂a

∂  (x − a)2 + (y − b)2 + z 2 = r 2 . ∂b

or, x − a = 0, and y − b = 0. Eliminate a and b from (2), z = ±r.

(6)

This result satisfies equation (4), but, unlike the particular solution, is not included in the complete internal (2). Such a solution of the differential equation is said to be a singular solution. Geometrically, the singular solution represents two plane surfaces touched by all the spheres represented by equation (2). The singular solution is thus the envelope of all the spheres represented by the complete integral. If AB (Fig. 79) represents a cross section of the xy-plane containing spheres of radius a, CD and EF are cross sections of the plane surfaces represented by the singular solution. If the one constant is some function of the other, say, a = b, (2) may be written (x − a)2 + (y − a)2 + z 2 = r 2 . 342

(7)

Differentiate with respect to a. We find a=

1 (x + y) . 2

Eliminate a from (7). The resulting equation x2 + y 2 + 2z 2 − 2xy = r 2 . is called a general integral of the equation. Geometrically, the general integral is the equation to the tubular envelope of a family of spheres of radius r and whose centres are along the line x = y. This line corresponds with the axis of the tube envelope. The general integral satisfies (4) and is also contained in the complete integral. Instead of taking a = b as the particular form of the function connecting a and b, we could have taken any other relation, say a = 12 . The envelope of the general integral would then be like a tube surrounding all the spheres of radius r whose centres were along the line x = 21 y. Had we put a2 − b2 = 1, the envelope would have been a tube whose axis was an hyperbola x2 − y 2 = 1. A particular solution is one particular surface selected from the double infinite series represented by the complete solution. A general integral is the envelope of one particular family of surfaces selected from those comprised in the complete integral A singular solution is the complete envelope of every surface included in the complete integral.* Theoretically an equation is not supposed to be solved completely until the complete integral, the general integral and the singular solution have been indicated. In the ideal case, the complete integral is first determined; the singular solution obtained by the elimination of arbitrary constants as indicated above; the general integral then determined by eliminating a and f (a). Practically, the complete integral is not always the direct object of attack. It is usually sufficient to deduce a number of particular solutions to satisfy the conditions of the problem and afterwards to so combine these solutions that the result will not only satisfy the given conditions but also the differential equation. * The study of Gibbs’ “Surfaces of Dissipated Energy,” “Surfaces of Dissociation,” “Surfaces of Chemical Equilibrium,” as well as “van der Waals’ Surfaces,” is the natural sequence of § 68, 126 and the present section. But to enlarge upon this subject would now cause a greater digression than is here convenient. Airy’s little book, in Elementary Treatise on Partial Differential Equations, will repay careful study in connection with the geometrical interpretation of the solutions of partial differential equations.

343

Of course, the complete integral of a differential equation applies to any physical process represented by the differential equation. This solution, however, may be so general as to be of little practical use. To represent any particular process, certain limitations called limiting conditions have to be introduced. These exclude certain forms of the general solution as impossible. See examples at the end of Chapter VIII.; also example (1) last set § 138, and elsewhere. The more important varieties of partial differential equations from the point of view of this work are the linear equations of the second and higher orders. § 139. The Solution of Partial Differential Equations of the First Order. For the ingenious general methods of Lagrange, Charpit, etc., the reader will have to consult the special textbooks, say, Forsyth’s A Treatise on Differential Equations (Macmillan & Co., 1888). Type I. The variables do not appear directly. The general form is, f (∂z/∂x, ∂z/∂y) = 0.

(I.)

The solution is z = ax + by + C, provided a and b satisfy the relation f (a, b) = 0, or b = f (a) . The complete integral is, therefore, z = ax + yf (a) + C.

(1)

EXAMPLES.–(1) Solve (∂z/∂x)2 + (∂z/∂y)2 = m2 . The solution is z = ax + by + C 0 , √ provided a2 + b2 = m2 . The solution is, therefore, ax + y m2 − a2 + C. For the general integral, put C = f (a). Eliminate a between the two equations, p ay + f 0 (a) = 0. z = ax + y m2 − a2 + f (a) ; and x − √ m2 − a 2

in the usual way. (2) Solve pq = 1. Ansr. z = ax + y/a + f (a). NOTE.–We shall sometimes write, for the sake of brevity, ∂z/∂x = p; ∂z/∂y = q.

(3) Solve a (p + q) = z. Sometimes, as here, when the variables do appear in the equation, the function of x, which occurs in the equation, may be associated with ∂z/∂x, or a function of y with ∂z/∂y, by a change in the variables. We may write the given equation ap/z + aq/z = 1. Put dz/z = dZ; dy/a = dY , dx/a = dX; hence, ∂Z/∂Y + ∂Z/∂X = 1, the form required.

344

Most modern texts use the phrase, “boundary conditions” rather than “limiting conditions”. KH

(4) Solve x2 p√2 + y 2 q 2 = z 2 . Put X = log x, Y = log y, Z = log z. Proceed as before. 2 Ansr. z = Cxa y 1−a .

If it is not possib1e to remove the dependent variable z in this way, the equation will possibly belong to the following class: Type II. The independent variables x and y are absent. The general form is, f (z, ∂z/∂x, ∂z/∂y) = 0. (II.) Assume as a trial solution, that ∂z/∂y = a · ∂z/∂x.

Let ∂z/∂x be some function of z obtained from II., say y = φ (z). Substitute these values in dz = p · dx + q · dy. We thus get an ordinary differential equation which can be readily integrated. dz = φ (z) · dz + aφ (z) · dy. Z ∴ x + ay = dz/φ (z) + C.

(2)

EXAMPLES.–(1) Solve p2 z + q 2 = 4. Here, p
2 a2 + z (dz/dx) = 4. a2 + z · dz/dx = 2, Z p
3/2
3 2 . Ansr. 2 a2 + z = 3 (x + ay + C) . a2 + z · dz = 12 a2 + z ∴ x+C = 2

2

(2) Solve p (1 + q) = q (z − a). Ansr. 4 (bz − ab − 1) = (x + by + C) .

If z does not appear directly in the equation, we may be able to refer the equation to the next type. Type III. z does not appear directly in the equation, but x and ∂z/∂x can be separated from y and ∂z/∂y. The leading type is f1 (x, ∂z/∂x) = f2 (y, ∂z/∂y) .

(III.)

Assume as a trial solution, that each member is equal to an arbitrary constant a, so that ∂z/∂x and ∂z/∂y can be obtained in the form, ∂z/∂x = φ1 (x, a) ; ∂z/∂y = φ2 (y, a) . dz = p · dx + q · dy. then assumes the form dz = φ1 (x, a) dx + φ2 (y, a) dy. EXAMPLES. –Solve the following equations: (1) q − p = x − y. Put ∂z/∂x + x = ∂z/∂y + y = a. Write ∂z/∂x = a − x, etc.; ∂z/∂y = a − y. Hence, z = − 21 (a − x)2 − 21 (a − y)2 + C. 3/2 3/2 (2) q 2 + p2 = x + y. Ansr. z = 32 (x + a) + 32 (y − a) + C. 2 2 2 (3) q = 2yp . Ansr. z = ax + a y + C.

345

(3)

←−

Type IV. Analogous to Clairault’s equation. The general type is z = p · x + q · y + f (p, q) .

(IV.)

The complete integral is z = ax + by + f (a, b)

(5)

EXAMPLES.–Solve the following equations: (1) z = px + qy + pq. z = −xy. p Ansr. z = ax + by + ab. Singular solution √ (2) z = px + qy + r 1 + p2 + q 2 . Ansr. z = ax + by + r 1 + a2 + b2 . Singular solution, x2 + y 2 + z 2 = r2 . The singular solution is, therefore, a sphere; r, of course, is a constant. √ √ (3) z = px + qy − n n pq. Ansr. z = ax + by − n n ab. Singular solution, 1/(2−n) z = (2 − n) (xy) .

§ 140. Partial Differential Equations of the nth Order. These are the most important equations that occur in physical mathematics. There are no general methods for their solution, and it is only possible to perform the integration in special cases. The greatest advances in this direction have been made with the linear equation. Before proceeding to this important equation, it appears convenient to solve some simpler types. EXAMPLES.–Integrate the following equations. ∂z ∂z ∂2z = a. If = p and = q. Integrate with regard to y and we get (1) ∂x∂y ∂x ∂y p = ay + f 0 (x). possible that f 0 (x) is a function of y. Integrate with respect R It is very 0 to x and z = {ay + f (x)} dx = axy + f1 (x) + f2 (y). ∂2z x (2) − = a. Ansr. z = 21 x2 log y + f1 (x) + f2 (y). ∂x∂y y R R  ∂2z ∂z (3) + f (x) = ψ (y). Ansr. z= e−yf (x) eyf (x) ψ (y)dy + f1 (x) dx+f2 (y). ∂x∂y ∂x

There are many points of analogy between the partial and the ordinary linear differential equations. Indeed, it may almost be said that every ordinary differential equation between two variables is analogous to a partial differential in the same class. The solution is in each case similar, but with these differences: First, the arbitrary constant of integration in the solution of an ordinary differential equation is replaced by a function of a variable or variables. Second, the exponential form, Cemx , of the solution of the ordinary ∂ linear differential equation assumes the form emx ∂y φ(y) .

346

See supplement page S-49 for the method of finding singular solutions to Clairault analog equations. KH

mx

∂

φ(y)

The expression, e ∂y , is known as the symbolic form of Taylor’s theorem. ∂ Having had considerable practice in the use of the symbol of operation D for ∂x , we ∂ may now use D0 to represent the operation ∂y . By Taylor’s theorem, φ (y + mx) = φ (y) + mx

∂φ (y) m2 x2 ∂ 2 φ (y) + ..., + · ∂y 2! ∂y 2

where x is regarded as constant.   m2 x2 ∂ 2 ∂ + . . . φ (y) . + ∴ φ (y + mx) = 1 + mx ∂y 2! ∂y 2 ∂ φ(y) mx ∂y

The term in brackets is clearly an exponential series (page 280) equivalent to e ∂ or, writing D0 for ∂y , 0

φ (y + mx) = emxD φ(y)

,

(1)

The general form of the linear equation is, A0

∂2z ∂2z ∂2z ∂z ∂z + A + A + A3 + A4 + A5 z = A, 1 2 2 2 ∂x ∂x∂y ∂y ∂x ∂x

(2)

where A1 , A2 ,. . . , A, may be constants, or functions of x and y. As with ordinary linear equations, Complete Solution = Particular Integral + Complementary function.

The complementary function is obtained by solving the left-hand side of equation (2), equated to zero. We may write (2) in symbolic form,
A0 D 2 + A1 DD 0 + A2 D20 + A3 D + A4 D 0 + A5 z = 0. (3) where D is written for stand

∂ ; ∂x

D 0 for

∂ ; ∂y

DD 0 for

∂2 . ∂x∂y

Sometimes we under-

F (D, D 0) z = 0.

(4)

in place of (2).

§ 141. Linear Partial Equations with Constant Coefficients. A. Homogeneous equations. Type: A0

∂2z ∂2z ∂2z + A + A = R, 1 2 ∂x2 ∂x∂y ∂y 2

(5)

where R is a function of x. To find the complementary function, put R = 0, and. instead of assuming, as a trial solution, that y = emx , as was the case with the ordinary equation, suppose that z = φ (y + mx) , (6) 347

is a trial solution. Differentiate (6), with respect to x and p, we thus obtain, ∂z = mf 0 (y + mx) ; ∂x

∂z = f 0 (y + mx) ; ∂y

∂2z = m2 f 00 (y + mx) ; ∂x2

∂2z = mf 00 (y + mx) ; ∂x∂y

∂ 2 z 00 f (y + mx) ; ∂y 2

Substitute these values in equation (5) equated to zero, and divide out the factor f 0 (y + mx). The auxillary equation, A0 m2 + A1 m + A2 = 0.

(7)

remains. If m is a root of this equation, f 0 (y + mx) = 0, is a part of the complementary function. If α and β are the roots of (7), then 0

0

z = eαxD φ1 (y) + eβxD φ2 (y) , as in (3), § 130. From (6), therefore,

z = f1 (y + αx) + f2 (y + βx)

(8) (9)

since α and β are the roots of the auxillary equation (7), we can write (5) in the form, (D + αD 0 ) (D + βD 0) = 0. (10) EXAMPLES.–Solve the following equations: ∂2z ∂ 2z (1) − = 0. Ansr. z = f1 (x + y) + f2 (y − x). ∂x2 ∂y 2 ∂2z ∂ 2z ∂2z − 4 = 0. Ansr. z = f1 (y − 2x) + f2 (y − 2x). + 4 (2) ∂x2 ∂x∂y ∂y 2 ∂2z ∂ 2z ∂ 2z − 2 2 = 0. Ansr. z = f1 (2y − x) + f2 (y + 2x). (3) 2 2 − 3 ∂x ∂x∂y ∂y 2 2 ∂ u ∂ u (4) = a2 2 . Ansr. u = f1 (at + x) + f2 (at − x). This most important equa∂t2 ∂x tion, sometimes called d’Alembert’s equation, represents the motion of vibrating strings, the law for small oscillations of air in narrow tubes (organ pipes), etc. We cannot say much about the undetermined functions f1 (at + x) and f2 (at − x) in the absence of data pertaining to some specific problem. Consider a vibrating harp string, where no force is applied after the string has once been put in motion. Let x = l denote the length of the string under a tension equal to the weight of a length L of the same kind of string. In order to avoid a root sign later on, a2 has been written in place of gL, where g represents the constant of gravitation. Further, let u represent the displacement of any part of the string we please, and let the ordinate of one end of the string be zero. Then, whatever value we assign to the time t, the limiting conditions are u = 0, when x = 0; and u = 0, when x = l. ∴ f1 (at) + f2 (at) = 0; f1 (at + l) + f2 (at − l) = 0, are solutions of dAlemberts equation. From the former, it follows that f1 (at)

must always be equal to

− f2 (at) ;

∴ f1 (at + l) − f1 (at − l) = 0.

348

←−

But at may have any value we please. In order to fix our ideas, suppose that at −l = q, ∴ at + l = q + 2l, where q has any value whatever. ∴ f1 (q + 2l) = f1 (q) . The physical meaning of this solution is that when q is increased or diminished by 2l, the value of the function remains unaltered. Hence, when at is increased by 2l, or, what is the same thing, when t is increased by 2l/a, the corresponding portions of the string will have the same displacement. In Other words, the string performs at least one complete vibration in the time 2l/a. Hence, we conclude that d’Alemberts equation represents a finite periodic motion, with a period of oscillation 2l/c. EXAMPLE.–Show that 1 2

{f1 (at + l) + f1 (at − l)} = 0.

is a solution of d’Alembert’s equation, and interpret the result. A further study of d’Alembert’s equation would require the introduction of Fourier’s series, Chapter VIII.

When two of the roots are equal, say α = β. We know that the solution of

(D − a)2 z = 0, z = emx (C1 x + C2 ) , § 130; by analogy, the solution of 2

(D − aD 0 ) z = 0; z = emx {xf1 (y) + f2 (y)} , z =xf1 (y + ax) + f2 (y + ax) .

or,

(11)

EXAMPLES.–Solve: ∂2z ∂2z ∂2z (1) ∂x 2 + 2 ∂x∂y + ∂y 2 = 0. Ansr. z = x f1 (y + x) + f2 (y + x).   2 3 (2) D3 − 3D2 D0 + DD0 + D0 z = 0.

Ansr. z = xf1 (y − x) + f2 (y − x) + f3 (y + x) .

The particular integral will be discussed after. B. Non-homogeneous equations. Type: ∂2z ∂2z ∂z ∂z ∂2z + A + A + A3 + A4 + A5 z = 0. (12) 1 2 2 2 ∂x ∂x∂y ∂y ∂x ∂y If the non-homogeneous equation can be separated into factors, the integral is the sum of the integrals corresponding to each symbolic factor, so that each factor of the form D − mD 0 , appears in the solution as a function of y + mx, and every factor of the form D − mD 0 − a, appears in the solution in the form z = eax f (y + mx). A0

EXAMPLES.– (1) Solve

∂2z ∂z ∂z ∂2z − + + = 0. ∂x2 ∂y 2 ∂x ∂y

Factors, (D + D0 ) (D − D0 + 1) z = 0. Ansr. z = f1 (y − x) + e−x f2 (y + x). ∂2z ∂2z ∂z ∂z (2) Solve − − − = 0. 2 ∂x ∂x∂y ∂x ∂y Factors, (D + 1) (D − D0 ) z = 0. Ansr. z = e−x f1 (y) + f2 (x − y).

349

←− ←−

It is, however, not often possible to represent the solutions of these equations in this manner. When this is so, it is customary to take the trial solution, x = eαx+βy , (13) and substitute for z in the given equation (12). Then, ∂z ∂2z ∂z = αz; = βz; = αβz; ∂x ∂y ∂x∂y ∂2z ∂2z 2 = α z; = β 2 z. 2 2 ∂x ∂y Equate the resulting auxillary equation to zero. We thus obtain
A0 α2 + A1 αβ + A2 β 2 + A3 α + A4 β + A5 z = 0.

(14)

This may be looked upon as containing a bracketed quadratic in α and β. For any value of β, we can find the corresponding value of α, or the value of α, for any assigned value of β. There is thus an infinite number of particular solutions of this differential equation. If u1 , u2 , u3 ,. . . , use particular solutions of any partial differential equation, each solution can be multiplied by an arbitrary constant and the resulting products are also solution of the equation. Similarly, it is not difficult to see that the sum of any member of particular solutions will also be a solution of the given equation. It is usually not very difficult to find particular solutions, even when the general solution cannot be obtained. The chief difficulty lies in the combining of the particular solutions in such a way, that the conditions of the problem under investigation are satisfied. Plenty of illustrations will be found at the end of the next chapter. If the above quadratic is solved for α in terms of β, and if the resulting f (α, β), is homogeneous, we shall have the roots in the form, α = m1 β; α = m2 β; , . . . , α = mn β. The equation will, therefore, be satisfied by any expression of the form, z = Σ Ceβ(y+mx) ,

(15)

where m has any value m1 , m2 ,. . . and C may have any value C1 , C2 ,. . . The symbol “Σ” indicates the sum of the infinite series, obtained by giving m and C all possible values. The above solution (15), may be put in a simpler form when β is a linear function of α, say, β = aα + b. This applies to equation (12). Again, we can sometimes solve the equation z = eαx+βy = 0, for α, in terms of β. In order to fix these ideas, let us proceed to the following examples. 350

Where m and C are not sets of discrete numbers, this sum becomes an integral taken dm where C is any integrable function of m. KH


EXAMPLES.–(1) Solve D2 − D0 z = 0. Hence α2 − β = 0. Put α = α = 2,. . . and we get the particular solutions

1 2,

α = 1,

1

e 4 (2x+y) , ex+y , e2x+4y , . . . Now the difference between any two terms of the form eαx+βy , is included in the above solution, it follows, therefore, that the first differential coefficient of eαx+βy , is also an integral, and, in the same way, the second, third and higher derivatives must be integrals. Thus we have the following particular solutions:– 2

Deαx+α

2

y

= (x + 2αy) eαx+α y .

2

y

2

y

= {(x + 2αy) + 2y} eαx+α y .

D2 eαx+α D3 eαx+α

2

= {(x + 2αy) + 6y (x + 2αy)} ,

etc.

If α = 0, we get the special case,

(2) Solve

∂2z ∂x2

−



z = C1 x + C2 x2 + 2y + C3 x3 + 6xy + . . .

∂2 z ∂y 2

∂z ∂z + 3 ∂y = 0. Put z = Ceαx+βy . Note − 3 ∂x

(α − β) (α + β + 3) = 0.

∴ β = α and β = 3 − α.

Hence z = Σ C1 eα(x+y) + e3y Σ C1 eα(x−y) ; z = f1 (y + x) + e3y f2 (y − x). ∂2z ∂x∂y 2

∂z ∂z + a ∂x + b ∂y + abz = 0. Ansr. z = e−ay f1 (x) + e−bx f2 (y).
(4) Solve D − D02 + D + 3D0 − 2 z = 0. Ansr. z = ex f1 (y − x) + e−2x f2 (y + x).

(3) Solve

§ 142. The Particular Integral of Linear Partial Equations. The following methods for finding the particular integral of homogeneous or non-homogeneous equations, are deduced by processes analogous to those employed for the particular integrals of the ordinary equations. The complementary function of the ordinary linear equation (D − m) z = 0, is, Cemx ; so, for the partial equation 0 (D − mD 0 ) z = 0, we have emxD φ(y) , or the equivalent φ (y + mx). This analogy extends to the particular integrals. In the former case, the particular integral of (D − m) z = R, is, z = (D − m)−1 R; while for F (D, D 0) z = R, we have z = F (D, D 0 )−1 R. Case 1 (General). When F (D, D) can be resolved into factors, so that, z = (D − mD 0 ) f (x, y) .

(16)

It is now necessary to find a value for this symbol (16). First show that Deαx R = (D + α) R, 351

by putting mD in place of α, and f (x, y), in place of R (Case 4, § 131). 1 1 0 0 ∴ f (x, y) = emxD e−mxD f (x, y) ; 0 0 D − mD D − mD 0 1 = emxD f (x, y − mx) . (17) D The value sought. The particular integral may, therefore, be found by the following series of operations: (1) Subtract mr from p in the f (x, y), to be operated upon. (2) Integrate the result with respect to dx. (3) Add mx to y, after the integration. If there is a succession of factors, the rule is to be applied to each one seriatim, beginning on the right. EXAMPLES.–Find particular integrals in the following examples. It is well to be careful about the signs of the different terms added and subtracted. It is particularly easy to err by want of attention to this. 1 1 ∂2z 2 ∂2 z (1) ∂x 2 − a ∂y 2 = xy. The particular integral is D−aD 0 · D+aD 0 xy. Now xy becomes x (y − ax). This, on integration with respect to dx, becomes 12 x2 y − 13 ax3 , and finally 1 1 1 2 3 2 x y − 2 ax − 3ax3 . Hence, 1 1 x2 y ax3 1 · xy = · + . D − aD0 D + aD0 D − aD0 2 6 Subtract −ax from y, for 12 x2 (y + ax) + 16 ax3 . Integrate and add −ax to the result. 1 3 result. 6 x y remains. This is the required
2 0 02 −1 (2) D + 3DD + 2D x + y. Ansr. 12 x2 y − 31 x3 .

Case 2 (Special). When R has the form f (ax + by). Multiply F (D, D 0 ) z by D n and get F (D, D 0 ) must be D m φ (D 0 /D) z = f (ax + by) . homogeneous in Operate on ax + by with D 0 and D respectively, powers

b D0 f (ax + by) = . D a As on page 313, the particular integral is 1 1 z= n· f (ax + by) . D φ (D 0 /D) ZZ Z 1 = . . . f (ax + by) dxn . φ (b/a) How to use this formula will appear from the examples.
EXAMPLES.–Find particular integrals in, (1) D2 + DD0 − 2D02 z = sin (x + 2y). The particular integral is ZZ 1 1 sin (x + 2y) = sin (x + 2y) dx2 ; D2 (1 + D0 /D − 2D02 /D2 ) 1+2−8 = 51 sin (x + 2y) .
(2) D2 + 5DD0 + 6D02 z = 1/ (y + 2x). Ansr.

352

1 20

{(y + 2x) log (y + 2x) − y − 2x}.

in order that 0 φ (D /D) will be a function of a single variable. KH Number of integrations is equal to the highest power of D in F (D, D 0 ). KH

The above process cannot be employed when F (D, D 0), or F (a, b) has the same form as R, because a vanishing factor then appears in the result. In such a case, use the above method for all factors which do not vanish when a is put for D, b for D. The solution is then completed by means of the formula: 1 f (y + mx) = xf (y + mx) . (18) D − mD 0 EXAMPLE.–Evaluate the particular integral in (D − D0 ) (D + 2D0 ) z = x + y.

For the first factor, use the above method and then 1 1 1 xD0 −xD0 = · (x + y) = e e (x + y) ; 0 D − D 3D 3D 1 xD0 1 1 xD0 1 1 = e y= e xy = x3 + x2 y. 3D D 3D 9 6

For a fully worked example of this type of equation, see the top of S-53 of the supplement. KH

Case 3 (Special). When R has the form of sin (ax + by), or cos (ax + by). Proceed. as on page 313, when 1 1 sin (ax + by) = sin (ax + by) , (19) 2 0 02 2 F (D , DD , D ) (−a , −ab, −b2 ) and in the same way for the cosine. EXAMPLES.–Find the particular integrals
(1) D2 + DD0 + D0 − 1 z = sin (x + 2y).

1 1 sin (x + 2y) = sin (x + 2y) D2 + DD0 + D0 − 1 −1 − 2 + D0 − 1 D0 + 4 1 1 = 02 sin (x + 2y) = − 20 (D0 + 4) sin (x + 2y) = − 10 {cos (x + 2y) + 2 sin (x + 2y)} . D − 16 ∴

(2) (D + DD0 − 2D0 ) z = sin (x − y) + sin (x + y). Find the particular integral for sin (x − y), then for sin (x + y). Add the two results together. Ansr. 12 sin (x − y) + 13 cos (x + y). For the anomalous case proceed as in § 131.

Case 4 (Special). When R has the form eax+by , proceed as directed on page 312, 1 1 eax+by = eax+by , (20) 0 F (D, D ) F (a, b) that is to say, put a for D and b for D 0 . EXAMPLES.–Find particular integrals
in the following: (1) D2 − DD0 − 2D02 + 2D + 2D0 z = e2x+3y .
−1 2x+3y 1 2x+3y Ansr. D2 − DD0 − 2D02 + 2D + 2D0 e = − 10 e . 0 0 mx+ny mx+ny (2) (DD + aD + bD + ab) z = e . Ansr. e / (m + a) (n + b).

If F (a, b) = 0, proceed as on page 313, z=

Fa0

1 1 eax+by , eax+by ; or, z = 0 (a, b) Fb (a, b) 353

(21)

←−

This example is more easily solved using the method of undetermined coefficients. See supplement S-53. KH

where Fa0 or Fb0 , denotes the first differential coefficient with respect to the subscript. The two results agree with each other.
EXAMPLE.–Solve D2 − D02 − 3D + 3D0 z = ex+2y .

Ansr. f1 (x + y) + e3y f2 (y − x) − yex+2y .

Case 5 (Special). When R has the form xr y s , where r and s are positive integers. Operate with F (D, D 0 )−1 on xr y s expanded in ascending powers of r and s. EXAMPLES.–(1) Find the particular integral in:
D2 + DD0 + D − 1 z = x2 y. o n
2

−1  = − 1+ D2 +DD0 +D0 + D2 +DD0 +D0 + . . . x2 y. = − 1− D2 +DD0 +D0
= − 1 + D2 + DD0 + D0 + 2D2 D0 x2 y = x2 y − 2y − 2x − x2 − 4.

The expansion is not usually carried higher than the highest power of the highest power in f (x, y).
−1 (2) Evaluate D2 − D02 − 3D + 3D0 xy. Ansr. 1 3 1 2 2 1 1 x − x − x − xy − x2 y. 18 9 27 9 6
2 2 02 (3) D − a D z = x.   ZZ D02 1 1 1 x dx2 = x3 . = 2 1 + a2 2 + . . . x = 2 x = D D D 6 −

See S-54 of the supplement for why the above statement fails to apply to example 2. KH

Case 6 (Special). When R has the form eax+by X, where X is a function of x or y. Use F (D, D 0)

−1 ax+by

e

X = eax+by F (D + a, D + b)−1 X,

(22)

derived as on page 315. EXAMPLE.–Find the particular integrals in 2

∂ 2 z/∂x2 − ∂z/∂y = xeax+a y . Ansr. e

ax+a2 y

2 1 1 1 x = eax+a y · D2 + 2aD − D0 2a D
2 = eax+a y 14 x2 /a − x/a2 .

 −1 D 1+ x 2a

§ 143. The Linear Partial Equation with Variable Coefficients. These may sometimes be solved by transforming them into a form with constants. E.g., ∂ r+s z (i.) Any term xr y s ∂x r ∂y s , may be reduced to the form with constant coefficients, by substituting u = log x, v = log y. EXAMPLES.–Solve the equations: ∂2z ∂z ∂z ∂2z +x = 0. This reduces to (1) x2 2 − y 2 2 − y ∂x ∂y ∂y ∂x

354

∂2 z ∂u2

−

∂2 z ∂v 2

= 0.

Use the method in case 5 on this page to apply the operator to the polynomial after you have applied case 6 to this example. KH

Hence the solution of this equation, z = φ1 (u + v) + φ2 (u − v), must be re-converted into the form in x and y, thus, z = f1 (xy) + f2 (x/y). ∂2z ∂2z ∂2z + 2 = 0. Ansr. z = f1 (y/x) + xf2 (y/x). (2) x2 2 + 2xy ∂x ∂x∂y ∂y R ∂2z ∂z (3) (x + y) −a = 0. Put ∂z = v·∂x. Ansr. z = f1 (y)+ (x + y)a f20 (x)·dx. ∂x∂y ∂x ∂ and (ii.) The transformation may be effected by substituting ϑ = x ∂x ∂ 0 ϑ = y ∂y , and treating the result as for constant coefficients.

EXAMPLES.–(1) Solve the first two examples of the preceding set in this way. ∂2z ∂z ∂2z ∂z ∂2z + y 2 2 − nx − ny + nz = 0. (2) x2 2 + 2xy ∂x ∂x∂y ∂y ∂x ∂y ∴ {ϑ (ϑ − 1) + 2ϑϑ0 + ϑ0 (ϑ0 − 1) − nϑ − nϑ0 + n} z = 0.

Ansr. z = xn f1 (y/x) + xf2 (y/x).

§ 144. The Integration of Differential Equations in Series. When a function can be developed in a series of converging terms, arranged in powers of the independent variable, an approximate value for the dependent variable can easily be obtained. The degree of approximation attained obviously depends on the number of terms of the series included in the calculation. The older mathematicians considered this an underhand way of getting at the solution but, for practical work, it is invaluable. As a matter of fact, solutions of the more advanced problems in physical mathematics are nearly always represented in the form of an abbreviated infinite series. Finite solutions are the exception rather than the rule. EXAMPLES.–(1) Evaluate the integral in f (x) = 0. Assume that f (x) can be developed in a converging series of ascending powers of x, that is to say, f (x) = a0 + a1 x + a2 x2 + a3 x3 + . . .

(1)

By integration Z

Z


a0 + a1 x + a2 x2 + . . . dx; Z Z Z = a0 dx + a1 x dx + a2 x2 dx + . . . ;

f (x) dx =

= a0 x + 21 a1 x2 + 31 a2 x3 + . . . ;
= x a0 + 12 a1 x + 31 a2 x2 + . . . + C.

(2)

(2) It is required to find the solution of dy/dx = y, in series. Assume that y = f (x), has the form (1) above, and substitute in the given equation. (a1 − a0 ) + (2a2 − a1 ) x + (3a3 − a2 ) x2 + · · · = 0.

355

(3)

This equation would be satisfied, if

a1 = a0 ; a2 = 21 a1 = 12 a0 ; a3 = 31 a2 = Hence,

1 a ;. . . 3! 0

y = a0 φ (x) ,

where φ (x) = 1 + x + 2!1 x2 + 3!1 x3 + · · · = ex . Put a for the arbitrary function, ∴ y = aex . That this is a complete solution, is proved by substitution in the original equation. Write the original equation in the form y = vφ (x) , where v is to be determined. Hence, dy φ (x) + v 0 {φ0 (x) − φ (x)} = 0, dx since φ (x) satisfies the original equation, dy = 0, or v is constant. dx For equations of higher degree, we must proceed a little differently. For example: dy d2 y −x − cy = x2 . (4) dx2 dx (i.) The complementary function. As a trial solution, put y = axm . The auxillary equation is m (m − 1) a0 xm−2 − (m + c) xm = 0. (5) This shows that the difference between the successive exponents of x in the assumed series, is −2. The required series is, therefore, (3) Solve

y = a0 xm + a1 x2m−2 + · · · + an−1 xm+2n−2 + an xm+2n ,

which is more conveniently written

y=

∞ X

an xm+2n .

(6)

0

In order to completely determine this series, we must know three things about it. Namely, the first term, the coefficients of x and the different powers of x that make up the series. Substitute (6) in (4), ∞ X (m + 2n) (m + 2n − 1) an xm+2n−2 − (m + 2n + c) an xm+2n = 0, (7) 0

where n, has all values from zero to infinity. If x is a solution of (4), the coefficient of xm+2n−2 must vanish with respect to m. Hence by equating the coefficient of xm+2n−2 to zero,* (m + 2n) (m + 2n − 1) an − (m + 2n − 2 + c) an−1 = 0. (8) if n = 0, m = 0, or m = 1. When n is greater than zero, m + 2n − 2 + c an = . (9) (m + 2n) (m + 2n − 1) This formula allows us to calculate the relation between the successive coefficients of x by giving n all integral values 1, 2, 3,. . . * If we take the other part of the auxillary a diverging series is obtained, useless for our purpose.

356

←−

First, suppose m = 0, then we can easily calculate from (9), a1 =

c c (c + 2) a1 = a0 ; . . . 3·4 4! c4 x2 + ... 1 + c + c (c + 2) 2! 4!

c a0 ; 1 · 2

∴ y 0 = a0

a2 =

(10)

Next put m = 1, and, to prevent confusion, write b, in (9), in place of a, c + 2n − 1 bn = bn−1 ; 2n (2n + 1) proceed exactly as before to find successively b1 , b2 , b3 . . .  x3 x5 ∴ y 00 = b0 x + (c + 1) + (c + 1) (c + 3) + ... (11) 3! 5! The complete solution of the equation, is the sum of series (10) and (11), or if ay1 = y 0 , by2 = y 00 , y = ay1 + by2 , which contains the two arbitrary constants a and b. (ii.) The particular integral. By the above procedure we obtain the complementary function. For the particular integral, we must follow a somewhat similar method. E.g., equate (8) to x2 instead of to zero. The coefficient of m − 2, in (5), becomes m−2

2

m (m − 1)a0 x =x ; ∴ m − 2 = 2 and m (m − 1) a0 = 1; ∴ m = 4; a0 = 12 . c + 2n + 2 an = an−1 . 2 (n + 2) (2n + 3)

From (9)

Substitute successive values of n = 1, 2, 3, . . . in the assumed expansion, and we obtain (particular integral) = a0 xm + a1 xm+2 + a2 xm+4 + . . . , where a0 , a1 , a2 , . . . and m have been determined. (4) Solve d2 y/dx2 + xy = 0.     1 1·4 6 2 2·5 7 Ansr. y = a 1 − x3 + x − . . . + x − x4 + x − ... . 3! 6! 4! 7! The so-called Riccati’s equation, dy + by 2 = cxn dx has attracted a lot of attention in the past. Otherwise it is of no particular interest here. It is easily reduced to a linear form of the second order. Its solution appears as a converging series, finite under certain conditions. Forsyth (l.c.) or Johnson (l.c.) must be consulted for fuller details. A detailed study of the more important series employed in physical mathematics follows naturally from this point. These are mentioned in the next section along with the titles of special textbooks devoted to their use.

§ 145. Harmonic Analysis. One of the most important equations in physical mathematics, is

∂2V ∂2V 1 ∂V ∂2V + + = . (1) 2 2 ∂x ∂y ∂z 2 κ ∂t It has practically the same form for problems on the conduction of heat, the motion of fluids, the diffusion of salts, the vibrations of elastic solids and

357

The m−2 = 2 equation comes from equating the exponents. KH

flexible strings, the theory of potential, electric currents and numberless other phenomena. x, y, z are the coordinates of a point in space, t denotes the time and V may denote temperature, concentration of a solution, electric and magnetic potential, the Newtonian potential due to an attracting mass, etc., κ is a constant. If the second member is zero, we have Laplace’s equation, if the second number is equated to 4πρ; where ρ is a function of x, y, z, the result is known as Poissons equation. ∂2V ∂2V ∂ 2V + + = 0, is Laplace’s equation. ∂x2 ∂y 2 ∂z 2 ∂2V ∂ 2V ∂2V + + = 4πρ, is Poisson’s equation. ∂x2 ∂y 2 ∂z 2 The first member is written ∇2 V by some writers, ∆2 V by others. The equation is often more convenient to use in polar coordinates, viz., ∇2 V =

∂2V 1 ∂2V 2 ∂V cot θ ∂V 1 + · + · + 2 · + , 2 2 2 ∂r r ∂θ r ∂r r ∂θ r sin2 θ

(2)

where the substitutions are indicated in (11), § 48.* Any homogeneous algebraic function of x, y, z, which satisfies equation (1), is said to be a solid spherical harmonic. These functions are chief used for finding the potential on the surface of a sphere, due to forces which are not circularly symmetrical.† Particular solutions of (1) give rise, under. special conditions, to the so-called surface spherical harmonics, tesseral harmomics and toroidal harmonics. The series   x2 x4 xn 1− 2 + − ... , 2n Γ (n + 1) 2 (n + 1) 24 · 2! (n + 1) (n + 2) is called a Cylindrical Harmonic or a Bessel’s function of the nth order. The symbol Jn (x) is used for it. The series is a particular solution of Bessel’s equation.   d2 y 1 dy n2 y = 0. + · + 1 − dx2 x dx x2 If n = 0, the series is symbolised by J0 (x) and called a Bessel’s function, of the zeroth order. These functions are employed in physical mathematics when dealing with certain problems connected with equation (1). Another particular solution is   1 1 x4 x2 + + ..., J0 (x) log x + 2 − 2 2 2 2 ·4 1 2 called a Bessel’s function of the second kind (of the zeroth order), symbolised by K0 (x). Similarly, the solution of Legendre’s equation 1 − x2 is the series 1−


d2 y dy − 2x + m (m + 1) y = 0, dx2 dx

m (m + 1) 2 m (m − 2) (m + 1) (m + 3) 4 x + x − ... 2! 4!

* This transformation is described in the regular textbooks. But possibly the reader can do it for himself. † A point is said to be circularly symmetrical, when its value is not affected by rotating it through an angle about the axis.

358

Modern texts almost universally use ∇2 V , which is spoken as “delsquared vee.” KH

See S-58 for development of the series expansion of J0 . KH

Most modern texts use Y rather than K for Bessel functions of the second kind, with K reserved for modified Bessel functions. KH

written, for brevity, Pm (x). This furnishes the so-called Surface Zonal Harmonics, Legendre’s coefficients, or Legendrians. Another particular solution, x−

(m − 1) (m + 2) 3 (m − 1) (m − 3) (m + 2) 5 x + x − ... 3! 5!

written Qm (x), gives rise to Surface Zonal Harmonics of the second kind. Both series are extensively employed in physical problems connected with equation (1). The equation x2 − b2




x2 − c2


dy 

d2 y +x x2 − b2 + x2 − c2 − m (m + 1) x2 − b2 + c2 p y = 0, 2 dx dx

called Lam´ e’s equation, has “series” solution which furnishes Lam´es functions or Ellipsoidal Harmonics, users in special problems connected with the ubiquitous equation ∇2 V +

1 dV κ dt

The so-called hypergeometric or Gauss’ series, 1+

ab a (a + 1) b (b + 1) 2 x+ x + ..., 1!c 2!c (c + 1)

appears as a solution of certain differential equations of the second order, say, x (1 − x)

dy d2 y + {c − (a + b + 1) x} − aby = 0, dx2 dx

(Gauss’ equation), where a, b, c, are constants. The application of these series to particular problems constitutes that branch of mathematics known as Harmonic Analysis. But we are getting beyond the scope of this work; for more practical details, the reader will have to take up some special work such as Byerly’s Fourier’s Series and Spherical Harmonics. Weber and Riemann’s Die Partiellen Differential-Gleichungen der Mathematischen Physik is the textbook for more advanced work. Gray and Mathews have A Treatise on Bessel’s Functions and their Application, to Physics (Macmillan & Co., 1895).

359

Supplemental Notes, including Worked Solutions to Selected Examples These notes by Karl Hahn are furnished under Creative Commons 3.0 license. See http://creativecommons.org/licenses/by/3.0/ for details. You may redistribute these notes freely or incorporate them into other works provided Karl Hahn is attributed as their author and the distributed copies reference Creative Commons 3.0 license. Page 283 (§ 117. Separable Equations), example 2: Problem: solve dv µ v = 0. + dx x2 µ 1 2 Separating variables: v dv + µ dx x2 = 0. Integrating: 2 v − x = C, from which the book’s answer follows immediately.

√ √ example 3: Solve 1 + x2 dy =
y·dx. Separating variables: dy/ y = dx/ 1 + x2 or √ equivalently dy/ y − dx/ 1 + x2 = 0. Integrating both sides yields the book answer.

example 4: Solve y − x · dy/dx = a (y + dy/dx). Gathering terms: (1 − a) y = (x + a) dy/dx. Multiply by dx and divide by y (x + 1): (1 − a) dx/ (x + 1) = dy/y. Integrate: (1 − a) log (x + 1) + C 0 = log y, which is the log 0 of the book answer if you allow that C = eC .

example 5: Word problem. Let a be the constant of proportionality. Since the charge is being dissipated, it time-derivative is negative. So we say dE dt = −aE. Separating variables: dE/E = −a dt. Integrating: log E = −at + C, or equivalently, E = E0 e−at , where E0 = eC . Page 284 (§ 117. Separable Equations continued), example 7: The book appears to be mistaken on this one. Solving the problem stated, it sets up as: dy x =− dx y Multiplying by y dx to separate variables yields y dy = −x dx or equivalently y dy + x dx = 0. Integrating gives the equation of a circle centered at the origin with an arbitrary radius. To get the book’s answer, it would have to say, “What curves have slope −y/x to the x-axis?” That sets up to y/dy + x/dx = 0 or equivalently log y + log x = C 0 , 0 which is the log of the book’s answer if you allow C = eC . Supplementary problem – the cooling of an object: This is a common problem assigned in classes on differential equations. An object cools at a rate that is proportional to the difference between its temperature and the ambient temperature. Suppose such an object starts at 100◦ C. After 5 minutes its temperature is 60◦ C and after 10 minutes its temperature is 50◦ C. What is its temperature after 20 minutes? The differential equation is dT = (Ta − T ) k, dt where Ta is the constant ambient temperature, and k is the rate of proportionality. Separating variables: dT = −k dt, T − Ta whose solution, when integrated, is log (T − Ta ) = −kt + C. Taking the antilog: T = Ta + ∆T0 e−kt ,

where ∆T0 = eC is the initial difference between temperature of the object and ambient temperature. Observe that there are three unknowns in the above, Ta , ∆T0 , and k. The three conditions given in the problem correspond to

S-1

100◦ C = Ta + ∆T0 , 60◦ C = Ta + ∆T0 u,

(a) (b)

50◦ C = Ta + ∆T0 u2 , where u = e−k×5

minutes

(c)

. Subtracting (a) from (b) and (a) from (c) eliminates Ta . −40◦ C = ∆T0 (u − 1) ,
−50◦ C = ∆T0 u2 − 1 .

(d)

Taking the quotient of these two eliminates ∆T0 . u−1 4 = 2 , 5 u −1 which leads to the quadratic, 4u2 − 5u + 1 = 0, hence u = 1 or u = 41 . The former is impossible because it leads to a zero denominator when you try to solve (d) for 140 ◦ ◦ ∆T0 . The latter yields ∆T0 = 160 3 C. Further back-substitution yields Ta = 3 C. At 1 −kt 4 t = 20 minutes you have e = u = 256 . Hence T (20 minutes) =

140 ◦ 3 C

+

160 ◦ 768 C

=

375 ◦ 8 C

= 46.875◦C.

Page 284 (§ 117. Equations with homogeneous powers), example 1: The book t + log (1 − t) + log x = C. does steps up to integrating the substituted equation to get 1−t y Back-substituting t = x ,  y 1 + log x = C, y + log 1 − 1− x x x + log (x − y) + log x − log x = C, x−y which, after you cancel the log x terms, becomes the log of the book’s answer. dy + y = 0. Substituting y = tx and example 2: The equation is equivalent to (y − x) dx dy dt = t + x dx dx   dt + tx = 0, (tx − x) t + x dx   dt (t − 1) t + x + t = 0, dx dt + t = 0, t2 − t + (t − 1) x dx Z Z (t − 1) dt dx + = C0, t2 x 1 log t + + log x = C 0 . t

Back-substituting t = y/x: x + log x = C 0 , y x log y = C 0 − , y

log y− log x +

0

which is the log of the book’s answer, if you let C = eC . dy − xy − y 2 = 0. Making the same substitution exmple 3: Convert the equation to x2 dx as in the last example:

S-2

  dt x t+x − x2 t − x2 t2 = 0, dx   dt − t − t2 = 0, t+x dx dt x − t2 = 0, dx dt dx − = 0, t2 x 1 − − log x = C 0 . t 2

Back-substituting and rearranging: x , y If you take the antilog of the above, you will see that it disagrees with the book’s answer. Yet if you take the differential of the above, you do indeed get back the original equation, so the error seem to be in the book. The final answer should be x = Ce−x/y , where 0 C = eC . log x = C 0 −

example 4: Dividing by dx then making the substitution and integrating   dt x2 + t2 x2 = 2x2 t t + x , dx   dt 1 + t2 = 2t t + x , dx dt 1 + t2 = 2t2 + 2tx , dx dt 2 1 − t = 2tx , dx 2t dt dx , = x 1 − t2
log x + log 1 − t2 = C 0 .

Back-substituting and simplifying

  y2 log x+ log 1 − 2 = C 0 , x
2 log x+ log x − y 2 − 2 log x = C 0 ,
log x2 − y 2 − log x = C 0 , x2 − y 2 = C, x

from which the book’s answer follows immediately. Page 285 (§ 117. Equations that can be made homogeneous in powers), example 2: This equation is exact, so using the method outlined in this paragraph is very much the long road to its solution. Using the method explained in § 119, you can solve this one on inspection. Nevertheless we shall do it the hard way here. We assume we have h and k such that x = v + h and y = w + k yields the equation, (2v − w) dv + (2w − v) dw = 0. The sequence to the solution to this (using the substitution, w = tv) is

S-3

dw (2v − w) + (2w − v) = 0, dv  dt = 0, (2v − tv) + (2tv − v) t + v dv   dt = 0, 2 − t + (2t − 1) t + v dv   dt 2 − t + 2t2 − t + (2t − 1) v = 0, dv   dt = 0, 2 − 2t + 2t2 + (2t − 1) v dv dv 1 (2t − 1) dt + = 0, v 2 t2 − t + 1
log v + 12 log t2 − t + 1 = C 0 ,

Back-substituting and simplifying log v +

1 2


log w2 /v 2 − w/v + 1 = C 0 ,
2 2 1 = C 0, 2 log w − wv + v w2 − wv + v 2 = C,

0

where C = e2C . Replacing v with x − h and w with y − k, this becomes our solution equation, (y − k)2 − (y − k) (x − h) + (x − h)2 = C, whose differential is 2 (y − k) dy − (y − k) dx − (x − h) dy + 2 (x − h) dx = 0.

For h and k to be such that this is equivalent to the original equation, (2y − x − 1) dy + (2x − y + 1) dx, you must have −2h − k = 1 from the dx terms and −2k − h = −1 from the dy terms. Solving yields h = −1 and k = 1. Replacing h and k with these values in the solution equation produces the book’s answer. Page 285 (§ 117. Equations where h and k cannot be established), second example 1: Substituting z = 2x+3y, you also have dz/dx = 2+3dy/dx or equivalently dy/dx = 31 (dz/dx − 2). dy + (2x + 3y − 1) = 0, dx   1 dz (z − 5) − 2 + z − 1 = 0, 3 dx   dz (z − 5) − 2 + 3z − 3 = 0, dx dz (z − 5) − 2z + 10 + 3z − 3 = 0, dx dz + z + 7 = 0, (z − 5) dx (z − 5) dz + dx = 0. z+7

(2x + 3y − 5)

Since

z−5 z+7

=1−

12 z+7 ,

integrating followed by back-substitution yields z − 12 log (z + 7) + x = C, 2x + 3y − 12 log (2x + 3y + 7) + x = C, 3x + 3y − 12 log (2x + 3y + 7) = C.

which, when divided by 3, is the book’s answer.

S-4

Page 285, second example 2: The substitution is again z = 2x + 3y, so again, dy/dx = 13 (dz/dx − 2). dy (3y + 2x + 4) − (4x + 6y + 5) = 0, dx   1 dz (z + 4) − (2z + 5) − 2 = 0, 3 dx   dz 3z + 12 − (2z + 5) − 2 = 0, dx dz 3z + 12 − (2z + 5) + 4z + 10 = 0, dx dz = 0, 7z + 22 − (2z + 5) dx (2z + 5) dz = 0. dx − 7z + 22

Since

2z+5 7z+22

=

2 7

−

9 1 7 7z+22 ,

integration and back-substitution is x − 72 z +

7x − 2 (2x + 3y) + 3x − 6y +

9 7 9 7

9 49

log (7z + 22) = C,

log (14x + 21y + 22) = C, log (14x + 21y + 22) = C.

The above disagrees with the book’s answer by a scalar of the terms outside the log. Yet taking the differential of the above does produce the original equation, so it appears that the book is incorrect. Page 288 (§ 118. Eliminating parameters), example 1: Divide by x, then differentiate twice y = a + bx, x y 1 dy = b, − x dx x2 1 d2 y 1 dy 1 dy y − 2 − + 2 3 = 0. x dx2 x dx x2 dx x Multiply by x3 , then gather like terms to arrive at the book’s answer. example 2: Divide by x and differentiate y2 = 4m, x 2y dy y2 − 2 = 0. x dx x Multiplying by x2 and dividing by y yields √ 2x · dy/dx − y = 0.pThis disagrees with the books answer. Yet if you substitute y = 4mx and dy/dx = m/x into this solution, it works. So the book appears to be in error. example y +y 00 y + y 00

3: = α cos x = −α cos x = 0

+β sin x −β sin x

S-5

example y0 −ay y 0 − ay

4: First eliminate α: = αaeax +βbebx = −αaeax −βaebx = β (b − a) ebx

Then use that result to y 00 − ay 0 −by 0 + aby 00 y − (a + b) y 0 + aby

eliminate β: = βb (b − a) ebx = −βb (b − a) ebx = 0

example 5: Divide by a − x and differentiate dx 1 = k, dt a − x  2 1 dx d2 x 1 + 2 = 0, dt2 a − x dt (a − x)  2 dx d2 x = 0. (a − x) 2 + dt dt

(a)

The general solution to the original equation is x = a + Ce−kt . The implication is that given arbitrary k and C, this function always satisfies (a). Observe that x (t) is, among other things, the position of the end of a relaxing spring that pulls on a damper, where the fully relaxed position is at x = a. In this application, our result indicates that the square of the spring-end’s velocity is equal to the product of its acceleration times its distance from the fully relaxed position, regardless of the spring constant or damping constant. Similar relationships exist for analogous systems, such as the current in an electric circuit having a resistor and inductor in series, or the temperature of an object cooling to ambient temperature. Page 290 (§ 119. Exact Equations) example 3: M = a2 y + x2 and N = b3 + a2 x. Hence ∂N ∂M = a2 and = a2 . ∂y ∂x example 4: M = sin y + y cos x and N = sin x + x cos y. Hence ∂M ∂N = cos y + cos x and = cos x + cos y. ∂y ∂x Page 291, example 3: M = a2 y + x2 and N = b3 + a2 x. Hence Z Z N dy = b3 y + a2 xy + h (x) . M dx = a2 xy + 31 x3 + g (y) and

Assuming the two integrals are equal, we infer g (y) = b3 y and h (x) = 13 x3 . Either way you end up with the book’s answer.
example 4: M = x2 − y 2 and N = −2xy. Hence Z Z M dx = 13 x3 − xy 2 + g (y) and N dy = −xy 2 + h (x) . Assuming the two integrals are equal, we infer g (y) = 0 and h (x) = 13 x3 . This gives the solution, 13 x3 − xy 2 = C, from which the book’s answer follows immediately.

S-6

Page 293 (§ 120. Finding integrating factors, Rule I) example 2: Multiplying this out and extracting factors of x dy and y dx, y 2 (y dx) − x2 (2y dx) + y 2 (2x dy) − x2 (x dy) = 0.

Rearranging according to the factors outside the parentheses,

x2 (−2y dx − x dy) + y 2 (y dx + 2x dy) = 0.

In this arrangement, α = 2, β = 0, m = −2, n = −1, α0 = 0, β 0 = 2, m0 = 1, and n0 = 2. Hence the equations for k and k 0 are −2k − 1 − 2 = k 0 − 1 and

− k − 1 = 2k 0 − 1 − 2.

Multiply the second by 2 and subtract −2k − 3 = k0 − 1 −2k − 2 = 4k 0 − 6 −1 = −3k 0 + 5 Hence k 0 = 2 and k = −2. Using the formulae for finding the exponents of x and y in the integrating factor: km − 1 − α = 4 − 1 − 2 = 1 and kn − 1 − β = 2 − 1 = 1. A quick check finds that using the primed symbols you get the same exponents, as expected. Recasting the original equation using the integrating factor, xy

xy 4 − 2y 2 x3 dx + 2x2 y 3 − x4 y dy = 0,

which is indeed exact. Z M dx = 12 x2 y 4 − 21 y 2 x4 + g (y)

and

Z

N dy = 12 x2 y 4 − 21 x4 y 2 + h (x) .

Clearly g (y) = 0 and h (x) = 0. The solution, x2 y 4 −x4 y 2 = C, that follows is equivalent to the book’s answer. Observe that in other such equations, it is not necessary for k or k 0 (or m, m0 , n, n0 , α, etc.) to be integers.

The original equation in this example, y 3 − 2yx2 dx + 2xy 2 − x3 dy = 0, is homogeneous in powers, so it can also be solved using the method introduced in § 117. Substituting y = tx and dy/dx = t + x dt/dx:  

dt t3 x3 − 2tx3 + 2t2 x3 − x3 t + x = 0, dx  

dt t3 − 2t + 2t2 − 1 t + x = 0, dx t3 − 2t dt +t+x = 0, 2t2 − 1 dx dt 3t3 − 3t +x = 0, 2 2t − 1 dx
2t2 − 1 dt dx + = 0, 3t3 − 3t x   dt dx t 1 + 2 + = 0, t t −1 3 x
log t log t2 − 1 + + log x = C, 3   26 y y 2 log + log − 1 + 6 log x = C, x x2
2 log y − 2 log x + log y 2 − x2 − 2 log x + 6 log x = C,
y 2 y 2 − x2 x2 = C,

which is also equivalent to the book’s answer.

S-7

Page 293 (§ 120. Finding integrating factors, Rule II) example 1: Note that by supposition, all equations susceptible to rule II are homogeneous in powers and therefore can also be solved by the method introduced in § 117. In this example (which is equivalent to page 284, example 4), M = x2 y + y 3 and N = −2xy 2 . Hence M x + N y = x3 y + xy 3 − 2xy 3 = x3 y − xy 3 .

If the reciprocal of the above is an integrating factor, as the book suggests, then 2xy 2 x2 y + y 3 dx − dy = 0 x3 y − xy 3 x3 y − xy 3 is exact. Cancelling common factors 2y x2 + y 2 dx − 2 dy = 0. x3 − xy 2 x − y2 Letting M 0 be the left-hand function of the above and N 0 be the right, then

x3 − xy 2 (2y) − x2 + y 2 (−2xy) ∂M 0 = 2 ∂y (x3 − xy 2 ) =

  3 3   2x3 y −  2xy + 2x3 y +  2xy 2

x2 (x2 − y 2 ) 0 ∂N −2x = (−2y) 2, 2 ∂x (x − y 2 )

=

4x3 y

2 2 x (x2 − y 2 ) 

,


−1 proving that the equation has indeed been made exact, hence µ = x3 y − xy 3 is an integrating factor of the original equation. Integrating Z Z M 0 dx = log x2 − y 2 − log x + g (y) and N 0 dy = log y 2 − x2 + h (x) . So g (y) = 0, h (x) = − log x, and the solution is x2 − y 2 = Cx.

Page 293, example 2: In this equation, M = x − ny and N = y. So the integrating factor is 1 −1 . µ = (M x + N y) = 2 x − nxy + y 2 This leads to the modified equation: y x − ny dx + 2 dy = 0. µM dx + µN dy = 2 2 x − nxy + y x − nxy + y 2 Letting M 0 = µM and N 0 = µN , we have
x2 − nxy + y 2 (−n) − (x − ny) (−nx + 2y) ∂M 0 , = 2 ∂y (x2 − nxy + y 2 )  − ny 2 +   + 2ny 2 2 2   −nx + nx − 2xy −  n2 xy n2 xy = . 2 (x2 − nxy + y 2 ) ∂N 0 −2xy + ny 2 = 2, ∂x (x2 − nxy + y 2 )
−1 Hence the modified equation is indeed exact and µ = R x2 − nxy + y 2 is an integrating factor for the original equation. Showing that M 0 dx = f (x, y) + g (y) and N 0 dy = f (x, y) + h (x), with the same f (x, y) for both, is messy and difficult for arbitrary n, but the reader is encouraged to try it for the easy case when n = 2. Use the substitution method in § 117 to show that
!1 y − n2 + α x 2α
= Cx y − n2 − α x is a solution for n 6= 2, where α2 = 14 n2 − 1.

S-8

Page 294 (§ 120. Rule III) example: From the problem we have M = (1 + xy) y and (1 − xy) x. Then 1 Mx − Ny =  xy xy  + x2 y 2 −   + x2 y 2 = 2x2 y 2 , hence µ = 2 2 . 2x y The modified equation is (1 − xy) x (1 + xy) y dx + = 0. 2 2 2x y 2x2 y 2 Again letting M 0 = µM and N 0 = µN , we find that
2x3 y − (1 + xy) 2x2 ∂M 0 = ∂y 4x4 y 2  3 3 2xy − 2x2 −  2x y = , 4 2 4x y
−2xy 3 − (1 − xy) 2y 2 ∂N 0 = ∂x 4x2 y 4   3   3 − 2y 2 +  −2xy 2xy  = , 2 4 4x y confirming that the modified equation is exact and that µ = 1/2x2 y 2 is an integrating factor. Z Z Z xy 1 dx + dx = − + 1 log x + g (y) , M 0 dx = 2 2 2x y 2x y 2xy 2 Z Z Z xy 1 dy 0 − dy = − − 1 log y + h (x) , N dy = 2xy 2 2xy 2 2xy 2 Clearly g (y) = − 21 log y and h (x) =

1 2

log x. So the solution is

1 − 1 log y + 12 log x = C. − (a) 2xy 2 Multiplying by 2, then moving everything except the log x term to the right-hand side and taking the exponential of both sides, 1

x = Cye xy . This solution differs from the book’s answer by the sign of the exponent. Yet if you take the differential of (a), you can readily return to the original equation. So it appears that the book is in error. Page 294 (§ 120. N = −2xy. Hence

Rule IV) example 1: In this equation, M = x2 + y 2 and

  1 ∂N 2 1 ∂M =− − (2y − (−2y)) = − . N ∂y ∂x 2xy x So the integrating factor is µ = e−2 log x = x−2 . The modified equation is   2y y2 dy = 0, µM dx + µN dy = 1 + 2 dx − x x which is easily verified as being exact. Integrating Z Z y2 y2 + g (y) and µN dy = − + g (x) . µM dx = x − x x 2

After you infer g and h, the solution is x− yx = C, from which the book’s answer follows immediately.

S-9

Page 294, example 3: In this equation, M = y 4 + 2y and N = xy 3 + 2y 4 − 4x. Hence   y 3 − 4 − 4y 3 − 2 1 ∂N ∂M −3y 3 − 6 3 = − = =− . M ∂x ∂y y 4 + 2y y 4 + 2y y So the integrating factor is µ = e−3 log y = y −3 . The modified equation is     4x 2 µM dx + µN dy = y + 2 dx + x + 2y − 3 dy = 0, y y which is easily verified as being exact. Integrating Z Z 2x 2x µM dx = xy + 2 + g (y) and µN dy = xy + y 2 + 2 + h (x) . y y

2 After you infer g and h, the solution is xy + 2x y 2 + y = C, from which the book’s answer follows when you multiply through by y 2 .

Page 297 (§ 122. 1st Order Linear Equations), example 1: The book already demonstrates the origin of the integrating factor, 1 µ= √ 1 + x2 Applying equation (2) of § 122 yields Z Z m dx y µm dx √ + C = = yµ = 3 + C. 2 2 1+x 1+x (1 + x2 ) 2 The remaining integral can be found by substituting tan u = x, which yields mx y √ = √ + C. 2 1+x 1 + x2 from which the book’s answer follows. Page 297, example 2: The symbology used in this problem is likely to be very confusing to students of electrical engineering. Virtually all modern text books use the symbol, i, to represent electrical current rather than C. So a modern text would di represent the equation to be solved here as E = iR + L dt . I will use today’s symbology in the solution of this example. E R di = . i + L dt L E Hence P = R L and Q = L . The integrating factor is µ=e Applying equation (2) of § 122, ie

Rt L

=

R

R L dt

Rt

=eL

E Rt e L dt + B, L

Z

E Rt e L + B, R Rt E + Be− L , i= R

Rt

ie L =

which is equivalent to the book’s answer.

S-10

Page 297, example 4: The equation is equivalent to dy y + = x2 dx x R So P = x1 and Q = x2 . The integrating factor is given by µ = e dx/x = x. So according to equation (2) of § 122 Z xy = x3 dx + C, xy = 41 x4 + C, C y = 14 x3 + , x

which agrees with the book’s answer. Page 298 (§ 122. Bernoulli’s and similar equations), example 1: Letting v = 1/y, as the book suggests, we have dv 1 dy =− 2 . dt y dx Dividing the equation by y 2 and multiplying by −1 yields the modified equation, 1 1 dy − = −1. − 2 y dx xy Substituting yields dv v − = −1. dx x This equation isRfirst-order linear, with P = −x−1 and Q = −1. Hence the integrating factor is µ = e− dx/x = x−1 . Applying equation (2) of § 122, Z dx v =− + C, x x v = − log x + C, x v = − log x + Cx, 1 = −x log x + Cx, y 1 = −xy log x + Cxy, which is equivalent to the book’s answer. Page 298, example 2: The original text had x sin2 x rather than x sin 2x as the second term. Although the book’s misprint is not obvious on first glance, it becomes obvious when you begin following the procedure the book gives for solving it. Using the original version of the equation in the book, you end up with a nonlinear equation when you make the substitution – an equation that is not susceptible by any of the methods covered so far. So my rendering of the text makes the correction I have already indicated. By trigonometric identity, the equation becomes dy + 2x sin (y) cos (y) = x3 cos2 y. dx Dividing by cos2 y, as the book suggests, and then applying more trig identities,
dy 1 + tan2 y + 2x tan y = x3 . dx
dy dv The book also suggests the substitution, v = tan y, whose derivative is dx = 1 + tan2 y dx . The equation becomes dv + 2xv = x3 . dx

S-11

Hence P = 2x and Q = x3 . The integrating factor is µ = e equation (2) of § 122 yields Z 2 2 ex v = x3 ex dx + C.

R

2x dx

2

= ex . Applying

To integrate the remaining integral, let u = x2 and du = 2x dx, then integrate by parts. Z 2 ex v = 21 ueu du + C, 2

ex v =

(u − 1) eu + C,
2 ex v = 12 x2 − 1 ex + C,
2 2 ex tan y = 21 x2 − 1 ex + C, 1 2

2

which is equivalent to the book’s answer. If you really needed y as a function of x, you  
−1 1 2 −x2 could rearrange this into y = tan . 2 x − 1 + Ce Comments on page 298, example 3: All the steps to solution are shown in the text, Z −Kx arriving at e e−Kx = −K dx + C. y x Handbook of Mathematical Functions, edited by Milton Abramowitz and Irene Stegun (U.S. Department of Commerce, National Bureau of Standards, Applied Mathematics Series · 55), defines the function Z ∞ −t e E1 (x) = dt t x in chapter 5 on the exponential integral function (by Walter Gautschi and William F. Cahill). According to this definition, example 3 is solved by

e−Kx = K E1 (Kx) + C. y This is the first example we have encountered so far in which the solution cannot be expressed in closed form. In this example there is no function that can be expressed as any finite combination of powers, roots, logs, exponentials, trig functions and inverse trig functions whose derivative is e−Kx /x. The E1 (x) function can only be approximated using power series or other convergent algorithms. In general, we say we have solved a differential equation if we can find a function solving the equation whose derivative is expressible in elementary functions, even if the solution itself cannot be. Page 299 (§ 123. Solution by Differentiation), case i, example 2: Substituting
the symbol, p, for dy/dx, the equation becomes xyp2 − x2 − y 2 p−xy = 0. This factors into (xp + y) (yp − x) = 0. This results in factored differential equation    dy y x dy = 0. + − dx x dx y dy y dy x Hence either =− or = . dx x dx y Both of these are separable resulting in dx dy =− or y dy = x dx. y x Integrating yields log y = log x + C or 12 y 2 = 21 x2 + C. The book’s answer follows from these. Page 299, case i, example 3: Substituting p for dy/dx, the equation is p2 −7p+12 = 0. This factors into (p − 4) (p − 3) = 0. Back-substitute and solve the two factors separately, and the book’s answer follows immediately.

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Page 299 case ii, comments: The book is overly brief on explaining this method of solution. The steps are to isolate either y or x and then differentiate with respect to the other independent variable. This eliminates the isolated variable. Now replace dy/dx or dx/dy with p and the second derivatives of y or x with the first derivative of p. In each of the examples, the resulting first order differential equation for p turns out to be separable. Solving that equation results in a solution containing an undetermined constant, C. Once you have that solution for p, you integrate that function to find the y or x function that is the solution to the original equation. This integration picks up an additional constant of integration, K. But unlike C, this second constant is not undetermined. In the earlier step where you took the derivative of the equation to eliminate one of the variables, you also lost a part of the equation. The constant, K, is the recovery of that lost part. So the final step of this method is to solve for K. To do this, substitute your solution function, including its added K, back into the original equation. You will find that not every value of K will make it work. Solving for the value or values of K that does make it work completes your solution. Page 299, case ii, example 1: The book shows the step of isolating y and taking the derivative of the resulting equation with respect to x. Making the substitution of p for dy/dx dp

1 dx √ , 2 p 1 dp (p − 1)dx = √ , 2 p 1 dp dx = , 2 p 23 − p 12 p=1+

which can be integrated by substituting u2 = p and 2u du = dp. The integral is shown in the book as x being a function of p, which then must be manipulated into p as a function of x.  √ p−1 1 x = log √ + log C, 2 p+1  √ p−1 2x , e =C √ p+1 √ √ e2x ( p + 1) = C ( p − 1) ,
√ p e2x − C = −C − e2x , r dy C + e2x , = dx C − e2x
2 C + e2x dy = (a) 2. dx (C − e2x ) This function can be integrated by substituting v = e2x and dv/(2v) = dx, then decomposing the resulting integrand into partial fractions.
2 Z C + e2x 2C (b) y= 2 dx = C − e2x + x + K. 2x (C − e ) We solve for K by substituting this expression for y into the original equation. To keep things tidy, we start by letting y = y0 + K. Then the substituted original equation is

S-13

d 2 (y0 + K) + 2x (y0 + K) = x2 + (y0 + K) , dx dy + 2xy0 + 2xK = x2 +y02 + 2Ky0 + K 2 . dx Now replace y0 with

2C C−e2x

+ x and dy/dx with the right-hand side of (a).
2   C + e2x 2C + x + 2Kx = + 2x 2 C − e2x (C − e2x )  2   2C 2C 2 x + + x + 2K + x + K 2. C − e2x C − e2x

Carefully multiplying all of this out and taking the cancellations 4Cx C 2 + 2Ce2x + e4x 2 =  + 2x +  2x + 2Kx 2  C −e (C − e2x ) x2 +

4C 2

(C −

2 e2x )

+

 4Cx 4KC 2  + K 2.  + 2Kx  + x +  C − e2x C − e2x

Now multiply what remains by C − e2x


2


2
C 2 + 2Ce2x + e4x = 4C 2 + 4KC C − e2x + K 2 C − e2x ,

C 2 + 2Ce2x + e4x = 4C 2 + 4KC 2 − 4KCe2x + K 2 C 2 − 2K 2 Ce2x + K 2 e4x .

(c)

Clearly e4x = K 2 e4x , hence K = 1 or K = −1. Trying these values of K on the remaining terms of (c) quickly reveals that only K = −1 works for the entire equation. Now replace K with −1 in (b) 2C y= + x − 1. C − e2x 2C C − e2x y= + x − , C − e2x C − e2x C + e2x y= + x, C − e2x which is the book’s answer. Page 299, case ii, example 2: First substitute p for dy/dx and solve for y xp2 − 2yp + ax = 0,   a x p+ = y. 2 p Taking the derivative using the product rule (remembering that p is the derivative of y)     a x a dp 1 p+ + 1− 2 = p. 2 p 2 p dx Before solving, subtract p from both sides and divide out the common factor     1 a x a dp −p + + 1− 2 = 0, 2 p 2 p dx     a a dp x p −1 + 2 + 1− 2 = 0, 2 p 2 p dx dp −p + x = 0, dx log p = log x + log C.

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Hence p = Cx and y=

Z

Z

p dx = C

x dx = 12 Cx2 + K.

We substitute this solution for y back into the original equation and solve for K.
2 x (Cx) − 2 12 Cx2 + K (Cx) + ax = 0,

C 2 x3 − C 2 x3 − 2CKx + ax = 0, a . K= 2C
The resulting solution, y = 12 Cx2 + a/C , does not exactly match the book’s answer, but this solution does work when substituted into the original equation, and the book’s does not. Furthermore, page 303 lists a solution for this same equation that is equivalent to the solution shown here (with the solution given on page 303, solve for y and replace C with 1/C 0 ). Page 299, case ii, example 3: The book instructs to solve for x and take the derivative with respect to y.  2 ! dy dy 2x , =y 1− dx dx   y dx dy , (d) − x= 2 dy dx   y 1 x= p− , 2 p where p is the derivative of x with respect to y. Taking the derivative with respect to y of the above eliminates x     1 y 1 dp 1 p− + 1+ 2 p= , 2 p 2 p dy     1 y 1 dp 1 p+ = 1+ 2 , 2 p 2 p dy dp p=y , dy log p = log y + C, dx p= = Cy. dy Integrating the last line of the above to get x as a function of y Z Z x = p dy = C y dy = 12 Cy 2 + K.

Substituting this solution into (d)

2 1 2 Cy

So the solution is x = 12 Cy 2 −

1 2C .

  1 Cy − , Cy 1 K =− . 2C

y +K = 2

Letting C 0 = 1/C

y2 − C 0, C0 y2 2x + C 0 = 0 , C 0 0 C (2x + C ) = y 2 . 2x =

which is equivalent to the book’s answer.

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Page 300, case iii, example 2: Both this example and the next succumb readily to the method of separation of variables (see § 117). Indeed using the method explained in § 123, case iii to do these seems almost an exercise in masochism, as it involves an order of magnitude more work. But to illustrate the method, here is example 2 done both ways. First by separation of variables. dy = dx, y + y1 y dy = dx, y2 + 1
2 1 2 log y + 1 = x +

1 2 2x

2

log C,

y + 1 = Ce .

from which the book’s answer follows immediately. But case iii would have you solve for y then differentiate. Letting p = dy/dx, 1 (a) p =y + , y py =y 2 + 1, y 2 − py + 1 = 0,   p y = 21 p ± p2 − 4 ,

p dp 1 dp dy =p= ± p dx . dx 2 dx 2 p2 − 4

This last equation is separable. Divide both sides by p, multiply by 2dx to get Z Z Z dp dp p , ± 2 dx = p p2 − 1   p 2x + log C = log p ± log p + p2 − 4 .

I will use the difference-solution of the ± and leave it up to the reader to try the sumsolution to confirm that it also arrives at the same final solution. p p Ce2x = , p + p2 − 4   p p + p2 − 4 Ce2x = p, p
Ce2x p2 − 4 = 1 − Ce2x p,   
C 2 e4x p2 , C 2 e4x p 2 − 4 = 1 − 2Ce2x +  
(b) −4C 2 e4x = 1 − 2Ce2x p2 , dy 2Ce2x √ =p= . 2x dx 2Ce − 1

(c)

√ Integrate (c) by substituting u = 2Ce2x and du 2Ce2x − 1 + K = y. 2u = dx to get √ Letting K = 0 (which you can confirm is correct by substituting y = 2Ce2x − 1 + K into the original equation), squaring both sides, and replacing 2C with C 0 yields an answer equivalent to the book’s. You could also have substituted (a) into (b), solved for y, and obtained the same result, but doing that algebra is more difficult than taking integral of (c).

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Page 300 (§ Clairaut’s equation), example 1: The equation given is already in Clairaut form. With p used to represent dy/dx, take the derivative of both sides: dp dp p
= p
+ x dx + 2p dx , which factors into dx = 0. (x + 2p) dp Either x = −2p or dp/dx = 0. If the former, then substitute p = −x/2 into the original equation. x2 x2 x2 y=− + =− , 2 4 4 from which one of the book’s answers follows. From dp/dx = 0 we have p = C and y = Cx + K. Substituting that into the original equation yields K = C 2 . Hence the other solution is y = Cx + C 2 . Page 300 (§ 124), example 2: To put this equation into Clairaut form, divide by p − 1 and then add px to both sides. p . (a) y = px + p−1 Taking the derivative dp dp 1 p
=p
+ x dx − (p − 1)2 dx , ! dp 1 . 0= x− 2 dx (p − 1) If the first factor is zero, then

1 √ + 1 = p. x

Substitute that into (a)   √1 + 1
2 √ √ √ √ 1 x = x+x+1+ x = x+2 x+1 = y = √ +1 x+ x+1 . 1 √ x x √ √ Hence y = ± ( x + 1). Regardless of which sign you choose from the ±, it still differs √ from the book’s answer by a sign. The solution, y = x+ 2 x + 1 works when substituted into the original equation. The book’s answer does not. The solution that results from C , the factor, dy/dx = 0, is again y = Cx + K. Substituting that into (a) yields K = C−1 from which the book’s other solution follows. Page 302 (§ 125. Singular solutions), Comments on p-discriminants and Cdiscriminants: My opinion is that Mellor does not expand enough on the relationship between p-discriminants, C-discriminant, and singular solutions. For one thing, he suggests that these discriminants exist only when the differential equation is a quadratic in p or when the solution is a quadratic in C. Chapter A2, section 10 of Ordinary Differential Equations and their Solutions by George M. Murphy (van Nostrand, 1960) provides a much broader method for finding discriminants. A p-discriminant occurs whenever x and y in the differential equation can conspire to cause p to have a double root. You find such double roots in the general case by taking the partial derivative of the equation with respect to p, solving the result for p, and substituting that expression for p back into the original equation. Mellor uses the example, xp2 − yp + a = 0, and, using the quadratic formula, arrives at the p-discriminant of y 2 = 4ax.

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Solving it Murphy’s way, we have ∂ (xp2 − yp + a) = 0, ∂p 2xp − y = 0, y p= . 2x Substituting p into the original equation,  y 2 y2 x − + a = 0, 2x 2x y2 + a = 0, − 4x y 2 = 4ax. Observe that Murphy’s method exploits the fact that a function has a double root only when both the function itself and its first derivative are simultaneously zero. The same method will also find the C-discriminant when applied to this equation’s general solution, xC 2 − yC + a = 0. The existence of a p-discriminant or a C-discriminant is a necessary but not sufficient condition for the existence of a singular solution. Both p-discriminants and Cdiscriminants of more complicated equations (an equation that is cubic in p for example) may be factorable, in which case there is a possibility for multiple singular solutions – that is each factor may be a singular solution. Murphy states without proof that if a factor of a p-discriminant is a singular solution, then there will be an identical factor of the C-discriminant, and vice versa. Mellor seems to suggest the same, but never states it outright. Both books are clear, however, that when a discriminant (or factor of a discriminant) is found, it must be tested by substituting it back into the original equation to see whether or not it is indeed a singular solution. Envelopment of singular solutions: As an example, the equation, px − y + p2 = 0, has as its general solution, y = Cx + C 2 . This solution has a C-discriminant of y = − 41 x2 . The figure shows how the family of curves specified by the general solution forms an envelope for the particular solution (general solution is shown with C = −0.2, −0.4, . . . − 2.0). To find an envelope point, we find the general solution for C and also for C + δC, then we find where those two lines intersect: y −y 0

= = =

(C + δC) x −Cx δCx

2

+ (C + δC) − C2 + 2C δC + δC 2

Graph showing lines, y = Cx + C 2 , enveloping the parabola, y = − 41 x2

Dividing out δC from the difference gives 0 = x + 2C + δC. Taking the limit as δC goes to zero gives x = −2C. Substituting
for x into the general solution gives y = −C 2 . So any point of the form, −2C, −C 2 , is a point on the envelope. Observe that the relationship between the expressions for x and y is precisely y = − 41 x2 , that is, the singular solution to the original equation. You could also have solved the equation, x = −2C for C and substituted into the general solution for C to obtain the same result. Observe also that solving for the intersection of neighboring lines in the family as we have done here is no different from Murphy’s method for finding the C-discriminant. By subtracting the neighbors from each other,

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dividing out δC, then taking the limit as δC goes to zero, we have effectively taken the partial derivative of the general solution with respect to C. After that we substituted that into the general solution, which is the same as the last step of Murphy’s method. This shows, at least for C-discriminants, how finding the conditions for C’s double root is identical to finding points on the curve that the general solution envelopes. Page 303 (Singular solutions), example 2: You arrive at the general solution by solving the differential equation for p, and integrating both sides. Use the substitution, u = x2 and du = 2x dx to integrate. Apply the constant of integration, C, to the y-side of the result, then square both sides. The book provides details for finding the p-discriminant and C-discriminant using the equal quadratic roots method. Applying 2 Murphy’s method to finding the p-discriminant, you have 4xp2 − (3x − a) = 0. Taking the partial with respect to p gives 8xp = 0. Hence either x = 0 or p = 0. Note that this p-discriminant is factorable. Substituting that into the original equation yields 2 (3x − a) = 0. So p-discriminants are x = 0 and x = 13 a. Using Murphy’s method to find the C discriminant, we take the partial of the general solution with respect to C and get 2(x + C) = 0, hence C = −x. Substituting into the general solution gives 2 0 = x (x − a) . Again the discriminant is factorable. x = 0 is the only common factor of both the p-discriminant and C-discriminant. It satisfies the original equation in that 2 if you divide the original equation by p2 (3x − a) and allow 1/p = dx/dy = 0 for the line defined by x = 0. Page 303 (Singular solutions), example 4: The general solution emerges as follows:
y 2 p 2 + 1 = a2 , ±a . y= p p2 + 1

Take the derivative of both sides to eliminate y: ±pa dp p= , 3 2 2 dx (p + 1) ±a dx = 3 dp, 2 (p + 1) 2 ±ap integrating yields x−C = p , p2 + 1 a2 p2 = (x − C)

2


p2 + 1 ,


0 = u 2 − a2 p 2 + u 2 ,

where u = x − C and du = dx. Hence

u2 , u 2 − a2 u . p = ±√ a2 − u 2

p2 = −

Integrating to establish y, q p 2 2 2 y = ± a − u + K = ± a2 − (x − C) + K.

2

Substituting back into the original equation yields K = 0. Hence y 2 = a2 − (x − C) , from which the book’s general solution follows. Rearrange the original equation into y 2 p2 + y 2 − a2 = 0. Using the quadratic method, the p-discriminant is y 2 (y 2 − a2 ) = 0. This discriminant factors into y = 0 or y = ±a. Rearrange the general solution into C 2 − 2xC + x2 + y 2 − a2 = 0. Using the quadratic

S-19

method, the C-discriminant is 4x2 = 4x2 + 4y 2 − 4a2 , or equivalently, y = ±a, which is in common with a p-discriminant factor. When y = ±a, then p = dy/dx = 0. So this factor works as a singular solution. Page 304 (§ 126. Trajectories), example 2: The transformation between Cartesian and polar coordinates is x = r cos θ and y = r sin θ. Taking the derivatives of these with respect to r,     dy dθ dθ dx and . = cos θ − sin θ r = sin θ + cos θ r dr dr dr dr By the chain rule we have   dy
sin θ + cos θ r dθ dy dr  dr
 = . (a) = dx dx cos θ − sin θ r dθ dr dr In Cartesian coordinates we find orthogonal trajectories by replacing dy/dx with −dx/dy.
Hence − cos θ + sin θ r dθ dx dr
. − = (b) dy sin θ + cos θ r dθ dr Replacing r dθ/dr with −dr/ (r dθ), the right-hand side of (a) becomes
dr sin θ − cos θ r1 dθ
. dr cos θ + sin θ 1r dθ Multiplying top and bottom by r dθ/dr yields the same expression as we got in (b) for −dx/dy, which completes the proof. Page 304 (§ 126. Trajectories), example 3: The curve family is defined by y 2 = 4ax. Solving this for a gives y2 = a. 4x Taking the derivative using the quotient rule, dy − 4y 2 8xy dx = 0. 16x2 Replacing dy/dx with −dx/dy and multiplying through by 4x2 , dx −2xy − y 2 = 0. dy An equivalent differential equation is 2x dx + y dy = 0. This is immediately integrable to x2 + 12 y 2 = C 2 , which is equivalent the book’s answer.

Page 304 (§ 126. Trajectories), example 4: Place two point charges of q and −q on the hor- Graph showing orthogonal trajec2 izontal axis symmetrically about the origin at a tories to y = 4ax distance of L apart, with the positive charge on the right. Allowing Lq to remain constant, we let L go to zero and q go to infinity to create an idealized dipole. Then the potential function, V , in polar coordinates is given by kLq cos θ , V = r2 where k is Coulomb’s constant. You could take the derivative with respect to r of this function and apply the transformation indicated in example 3 above, and you would arrive at the correct trajectory. You are encouraged to try that. You would find that, in the end, you would have to take the exponential of a function expressed in logs. So

S-20

here I will pass into logs before taking the derivative and solve it that way. The above equation becomes log V = log (kLq) + log (cos θ) − 2 log r. Taking the derivative with respect to r, dθ 2 − , 0 = − tan θ dr  r  dθ − 2, 0 = − tan θ r dr Now make the transformation from example 3:   1 dr − 2, 0 = tan θ r dθ dr 0= − 2 cot θ dθ, r dr 0 = 2 cot θ dθ − . r Integrating the above, log C = 2 log (sin θ) − log r, C=

sin2 θ . r

Equipotentials and trajectories of an ideal dipole. Page 307 (§ 128. Linear equations, when a particular solution is known), derivation of order reduction: We begin with the assumption that v is a known solution to the equation, dy d2 y +P + Qy = 0. dx2 dx where P and Q are functions of x. We let y = uv. Then, by the product rule, dy dv du =u +v dx dx dx

and

d2 v du dv d2 y d2 u = u + 2 . + v dx2 dx2 dx dx dx2

Substituting   du dv d2 u dv du d2 v + 2 + v + P u + v + Quv = 0. dx2 dx dx dx2 dx dx But because v is assumed to solve the original equation, u

(a)

dv dv d2 v d2 v + P + Pu + Qv = 0 = u + Quv. (b) 2 2 dx dx dx dx Removing from (a) the terms that occur to the right of the second equal in (b), what   remains is d2 u dv du v 2 + 2 + Pv = 0. dx dx dx

Page 307 (§ 128. Linear equations), example 1: y = eax is assumed to be a solution to d2 y d2 y 2 = a y, or equivalently, − a2 y = 0. dx2 dx2 Hence P = 0, Q = −a2 and v = eax . Substituting these into the formula on page 307, du + 2 log (eax ) = C, log dx du = Ce−2ax , dx

S-21

C −2ax e + C2 , −2a C y = uv = − e−ax + C2 eax . 2a u=

C yields the book’s answer. Letting C1 = − 2a

Page 307 (§ 128. Linear equations), example 2: y = x is assumed to be a solution to
d2 y dy dy x d2 y y 1 − x2 −x − = 0. + y = 0, or equivalently, + 2 dx dx dx2 1 − x2 dx 1 − x2
R 1 −x P dx = 21 log 1 − x2 . Using Hence P = 1−x 2 , Q = 1−x2 , and v = x. We also have the formula on page 307,
du + 2 log x + 12 log 1 − x2 = log C, log dx du C , = √ 2 dx x 1 − x2 √ C 1 − x2 + C2 . u=− x √ Because v = x, we have y = uv = ux = −C 1 − x2 + C2 x. Letting C1 = −C gives book’s answer. Page 308 (§ 129. Linear equations with constant coefficients), example 2: The auxillary equation is D2 − m2 = (D + m) (D − m) = 0, which has roots at ±m. The book’s answer follows. Page 308 (§ 129. Linear equations with constant coefficients), example 3: The auxillary equation is D2 + 4D + 3 = (D + 3) (D + 1) = 0, which has roots at −3 and −1. The book’s answer is missing minus signs in the exponents. Page 308 (§ 129. Linear equations with constant coefficients), case 2 comments: Rather than using the Maclaurin series to establish the general solution for the case of a double root, you can also use the formula on page 307 that establishes a general solution if a particular solution is known. The general case of a second order linear equation with constant coefficients that has a double root is dy d2 y − 2a + a2 y = 0, 2 dx dx whose auxillary equation has its double root at a. Because a is a root, we know that yR = eax is a particular solution. So P = −2a, Q = a2 and v = eax . We also have P dx = −2ax. Applying the formula du + 2 log (eax ) − 2ax = log C1 , dx du −  = log C1 , + 2ax 2ax log dx du = log C1 , log dx du = C1 , dx u =C1 x + C2 , y = uv = (C1 x + C2 ) eax . log

S-22

Page 309 (§ 129. linear equations with constant coefficients, case 3, nonreal roots) example 1: All that’s needed is to prove that eαx = cosh αx + sinh αx. By definition eαx − e−αx eαx + e−αx and sinh αx = , cosh αx = 2 2   + eαx −    hence e−αx e−αx eαx +  cosh αx + sinh αx = = eαx . 2 Page 309 (§ 129. linear equations with constant coefficients, case 3, nonreal √ roots) example 2: The auxillary equation is D2 + D + 1 = 0. Its roots are − 21 ± ι 23 . √ √ This means that y = C e(−1+ι 3)x/2 + C e(−1−ι 3)x/2 is the general solution. To turn 2

1

this into a real-valued function, let C1 = 12 (A − ιB) and C2 = conjugates of each other. Then √ √ y = 1 (A − ιB) e(−1+ι 3)x/2 + 1 (A + ιB) e(−1−ι 3)x/2 , 2

√

2

√

y = 12 (A − ιB) e−x/2 eι 3x/2 + 21 (A + ιB) e−x/2 e−ι 3x/2 , and √ √ !     3x 3x A B −x/2 B −x/2 A cos e + e −ι + ι sin +ι y= 2 2 2 2 2 2 √ √ √  3x B −x/2 A −x/2 A −x/2 3x 3x     cos +ι  e  sin − ι e cos + y= e 2 2 2 √2 √2 √2  3x 3x 3x A −x/2 A −x/2 B e cos −ι  e  sin + ι e−x/2 cos +    2 2 2 2 √ 2 √2 3x 3x + Be−x/2 sin . y =Ae−x/2 cos 2 2

1 2

(A + ιB) be complex

by Euler’s formula, √ √ ! 3x 3x cos , −ι sin 2 2 √ B −x/2 3x e sin + 2 2 √ 3x B −x/2 e sin , 2 2

Page 309 (§ 129. linear equations with constant coefficients, case
3, nonreal roots) example 4: The auxillary equation factors into (D − 1) D2 + 1 = 0. This has roots at 1 and ±ι. The book’s answer follow immediately from these roots. Page 311 (§ 130. Finding particular solutions) case 1, example 1: The illustration has you find the particular integral, p (x), by integrating (see equation (4) on page 311) Z Z Z Z p (x) = e3x e−3x e4x dx − e2x e−2x e4x dx = e3x ex dx − e2x e2x dx. The result is p (x) = e4x − 12 e4x , which, when simplified and combined with the complementary solution, yields the book’s answer.

Page 311 (§ 130. Finding particular solutions) case 1, example 2: Auxillary equation factors into (D − 3) (D − 1) = 0. The decomposition into partial fractions is 1 1 1 = 2 − 2 . (D − 3) (D − 1) D−3 D−1

Integrating using equation (4) on page 311, Z Z Z Z p (x) = 21 e3x 2e−3x e3x dx − 12 ex 2e−x e3x dx = e3x dx − ex e2x dx.

This yields p (x) = xe3x − 12 e3x . The second summand is linearly dependent upon complementary solution, so we can eliminate it from the particular solution, leaving p (x) = xe3x . Combining that with the complementary solution, C1 ex + C2 e3x , yields the book’s answer. Page 311 (§ 130. Finding particular solutions) case 2, example 2: In this case, R = ex . In problems of finding particular solutions, the expression in the Rposition is commonly referred to as the forcing function. When the forcing function is

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an exponential that is linearly independent of the complementary solution, there is an easy way to find the particular solution. When these conditions exist, the particular solution must be a constant multiple of the forcing function. So we let p (x) = λR. In this case that means p (x) = λex . We then substitute that into the original equation
and solve for λ. In this case, where the original equation is D2 − 4D + 4 y = ex , x x     λex −  4λe + 4λe = ex . Hence λ = 1 and p (x) = ex . So the general solution is y = C1 e2x + C2 xe2x + ex .

This method also works when the forcing function is the sum of exponentials, provided that none of the exponentials in that sum are linearly dependent
upon the complementary solution. So, for example, if we were to solve D2 − 4D + 4 y = e3x − 2e−3x , then p(x) = λ1 e3x + λ2 e−3x . Substituting into the original equation, 9λ1 e3x + 9λ2 e−3x − 12λ1 e3x + 12λ2 e−3x + 4λ1 e3x + 4λ2 e−3x = e3x − 2e−3x .

By classifying terms, this separates into

9λ1 − 12λ1 + 4λ1 = λ1 = 1 and 9λ2 + 12λ2 + 4λ2 = 25λ2 = −2.

2 Hence λ1 = 1 and λ2 = − 25 and p (x) = e3x −

2 −3x . 25 e

The procedures shown above for finding particular solutions for both R = ex and R = e3x − 2e−3x are examples of the method of undetermined coefficients. The principle of superposition: In the example above, we had a forcing function of R = R1 + R2 , where R1 = e3x and R2 = e−3x . If you solve the particular integral for just R1 by itself to get p1 (x) and for just R2 by itself to get p2 (x), you will find that p (x) = p1 (x) + p2 (x) is the particular integral we got for R = R1 + R2 . This is the principle of superposition. It asserts that whenever the forcing function is a sum of functions, you can find the particular integral for each summand separately, and the particular integral for the sum will be the sum of those particular integrals. In addition, for any constant scalar multiplier, α, the particular integral for a forcing function, αR, will be αp (x), where p (x) is the particular integral for R by itself. The principle of superposition of forcing functions and the assertion on scalar multiples of forcing functions are both generally true for all linear differential equations. Page 312 (§ 130. Finding particular solutions) case 3, example 1: This example too can be done using the method of undetermined coefficients. We observe that if the forcing function, R, is a polynomial of degree n, then the particular integral must also be a polynomial of degree n. In the example, n = 2. So we let p (x) = λ2 x2 + λ1 x + λ0 and substitute into into the original equation.

D2 − 4D + 4 λ2 x2 + λ1 x + λ0 = 2λ2 − 8λ2 x + 4λ2 x2 − 4λ1 + 4λ1 x + 4λ0 = x2 . Separating terms according to powers of x, 4λ2 x2 − 8λ2 x 2λ2

+ −

4λ1 x 4λ1

+ 4λ0

= = =

x2 , 0, 0.

Dividing out the powers of x from the first two lines and solving the linear system yields λ0 = 38 , λ1 = 12 , and λ2 = 41 , which is the same as the book’s answer. Page 312 (§ 130. Finding particular solutions) case 3, example 2: This one can be done by inspection using the method of undetermined coefficients. The forcing function, R, is the first degree polynomial, 2 + 5x. Since the second derivative of any first degree polynomial is zero, then if p (x) is the particular integral, substituting it into the original equation gives −p (x) = 2 + 5x, or equivalently, p (x) = −2 − 5x. Summing that particular solution with the complementary solution yields the book’s answer.

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Were you to do it by the method in the book, you would first resolve the reciprocal operator into partial fractions. 1 1 1 1 1 1 = = − . 2 D −1 (D + 1) (D − 1) 2D−1 2D+1 Hence the integral problem to be solved is Z Z −x 1 −x 1 x (2 + 5x) e dx − 2 e (2 + 5x) ex dx, p (x) = 2 e

which, after your expending a lot more work than you would using the first method, yields the same p (x).

Page 312 (§ 130. Particular solutions) case 4, clarification of notation: The first line of (5) on page 312 would be clearer as n

Dn (eax X) = eax (D + a) (X) , so as to show the extent of application of the differential operators. The final result of (5) would be clearer as D−n (eax X) = eax (D + a)

−n

(X) .

This is simply assertion that the first formula works even when n is negative. Page 312 (§ 130. Particular solutions) case 4, example 2: By (5) on page 312, (D − 1)−1 (ex log x) = eax (D + 1 − 1)−1 (log x) = eax D−1 (log x) .

The D−1 operator is simply the taking of the antiderivative. The antiderivative of log x is x (log (x) − 1) = x log (x/e). The book’s answer follows from this. Page 313 (§ 130. Particular solutions) case 4, example 1: By (5) on page 312,
−3 −3 (D + 1) e−x = e−x (D − 1 + 1) · 1 = e−x D−3 · 1.

The D−3 operator tells us to take the antiderivative of 1 three times. The first time gives x, the second, 12 x2 , and the third, 61 x3 . The book’s answer follows from that. Page 313 (§ 130. Particular solutions) case 4, example 2: First factor the

operator: D3 − 1 = (D − 1) D2 + D + 1 . By (5) on page 312,
−1
−1 x −1 x. (D − 1) xe = ex D−1 D2 + 3D + 3 D2 + D + 1
−1 Applying D−1 to x gives 21 x2 . We then apply D2 + 3D + 3 to 21 x2 . If p (x) is the function that solves this, then
D2 + 3D + 3 (p) = 21 x2 . (a)

We can apply the method of undetermined coefficients to this. Let p = λ2 x2 + λ1 x + λ0 . Applying the differential operator in (a) to p, 2λ2 + 6λ2 x + 3λ1 + 3λ2 x2 + 3λ1 x + λ0 = 12 x2 . Sorting terms by powers of x gives the linear system, 3λ2 x2 6λ2 x 2λ2

+ +

3λ1 x 3λ1 +

3λ0

= = =

1 2 2x ,

0, 0,


yielding λ2 = 61 , λ1 = − 31 , and λ0 = 92 . Hence p (x) = 16 x2 − 31 x + 29 ex . The book’s answer is missing the constant term inside the parentheses. You are invited, though, to demonstrate for yourself that
d3 p = 61 x2 + 32 x + 29 ex , 3 dx
which means that D3 − 1 p = xex , as the problem requires. The book’s answer does not, however, solve this equation.

S-25

Page 313 (§ 130. Particular solutions) case 5, example 1: This one factors
−1 −1 into p (x) = D2 + 1 (D + 1) sin 2x. Since n = 2, we have −n2 + 1 = −3. −1 So according to (9) on page 313, we have p (x) = − 13 (D + 1) sin 2x, or equiva1 lently, (D +
1) p (x) = − 3 sin 2x. When you multiply both sides by (D − 1), you get D2 − 1 p (x) = − 13 (D − 1) sin 2x. On the left we replace D2 with −n2 to get −5p (x) = − 13 (D − 1) sin 2x. Now apply the remaining differential operator to sin 2x and divide both sides by −5 to get the book’s answer. Page 314 (§ 130. Particular solutions) case 5, example 2: In the D-form, this
equation is D2 − k 2 y = cos mx. Clearly the roots for the formation of the complementary solution are ±k. We set up
−1 cos mx. p (x) = D2 − k 2

Here n = m, so we replace D2 with −m2 , so according to (10) on page 313 1 cos mx. p (x) = 2 −m − k 2 The book’s answer follows from this.

Page 314 (§ 130. Particular solutions) case 5, example 3: Setting this up in
D-form, D2 + mD + n2 y = a sin nt. So for the particular solution, p (x), we have
−1 a sin nt. p (x) = D2 + mD + n2

Replacing D2 with −n2 (in accordance with (9) on page 313) results in a cancellation inside the parentheses to give p (x) =

−1 1 a sin nt. mD

Again the D−1 operator is simply taking the antiderivative. Taking the antiderivative of sin nt gives − n1 cos nt. The book’s answer follows from that and from the given information that the roots for forming the complementary solution are α and β. In many engineering books, this equation is given in the form of dy d2 y + 2ζω + ω 2 = a sin ωt. dt2 dt Here ω is known as the resonant frequency of the system and ζ is known as the damping factor. The reciprocal of 2ζ is known as the Q-factor. You are encouraged to solve this problem in this form as well. What happens to the roots of the auxillary equation when 0 < ζ < 1? How does that affect the complementary solution? As ζ approaches zero, how does this affect the particular solution? This equation represents what engineers call a resonant system. Page 314 (§ 130. Particular solutions) case 5, example 1: That − 12 x cos x
−1 results from D2 + 1 sin x can be proved in the same way as is shown on page 314 for the identical operator on cos x. But with a little a priori knowledge, we can apply the method
of undetermined coefficients to this too. The a priori knowledge is that if D2 + n2 is a factor of the operator, and a sin nx + b cos nx is the forcing function, then the particular integral will be in the form of p (x) = λs x sin nx + λc x cos nx. In this case, a = 1, b = 0 and n = 1.
D2 + 1 (λs x sin x + λc x cos x) = sin x. Hence

    x sin x) + λc (−2 sin x −  x sin x + x cos x + λs (2 cos x −  x cos x) = sin x.

Clearly λs = 0 and λc = − 21 . The book’s answer follows from that.

S-26

See page S-61 for further development of resonant systems.


Similarly you can find that for arbitrary n, where D2 + n2 is a factor and sin nx is the forcing function,   (
(   (( (( n2 x cos x + n2 x cos x = sin nx. n2( x sin nx +λc −2n sin x −  n2( x sin nx + ( λs 2n cos nx − (

1 . In this case, λs = 0 and λc = − 2n

Page 314
(§ 130. Particular solutions) case 5, example
4: In the D-form, this is D3 − 1 y = x sin x, which factors into (D − 1) D2 + D + 1 y = x sin x. The roots of √ the auxillary equation are 1 and − 12 ±ι 12 3. These roots account for the complementary solution in accordance with case 3 on page 309. The book offers no satisfactory strategy for solving the particular integral on this one until case 6, so this may be regarded as a challenge problem. Here are two approaches you might take. The first is to observe that according to Euler’s formula   ιx
−1
−1 e − e−ιx 3 3 , p (x) = D − 1 x sin x = D − 3 x 2ι √ where ι = −1. By the principle of superposition    
−1 e−ιx
−1 eιx − D3 − 1 . x p (x) = D3 − 1 x 2ι 2ι By case 4 on page 312, this is the same as n
−1 x o
−1 x o n −ιx D3 − ι3D2 − 3D + ι − 1 − e . p (x) = eιx D3 + ι3D2 − 3D − ι − 1 2ι 2ι So if p (x) = p1 (x) − p2 (x) where
−1 x p1 (x) =eιx D3 + ι3D2 − 3D − ι − 1 2ι and
−1 x , p2 (x) =e−ιx D3 − ι3D2 − 3D + ι − 1 2ι

then we can solve for p1 and p2 separately using the method of undetermined coefficients, and then recombine the solutions to form p. If p1 = eιx (λ0 + λ1 x), then we need only concern ourselves with the (−3D − ι − 1) part of the differential operator, as higher derivatives of (λ0 + λ1 x) are zero. So for p1 we must solve x (−3D − ι − 1) (λ0 + λ1 x) = 2ι for λ0 and λ1 . Expanding this,

1 4

(−ι − 1) λ1 x −3 λ1 +

Hence λ1 = (1 + ι) and λ0 = methodology we also solve

− 43 ,

(−ι − 1) λ0

x = 2ι , = 0.

from which we can write p1 (x). Using the same

x . 2ι The result is λ3 = 41 (−1 + ι) and λ2 = 34 . This allows us to write p2 (x) as well as p1 (x). (−3D + ι − 1) (λ2 + λ3 x) =

p (x) = p1 (x) − p2 (x) = 14 eιx {−3 + (1 + ι) x} − 14 e−ιx {3 + (−1 + ι) x} ,


p (x) = − 34 eιx + e−ιx + 14 eιx + e−ιx x + 41 eιx − e−ιx ιx, and by Euler’s formula p (x) = − 32 cos x + 21 x cos x − 21 x sin x, which is the same as the book’s answer.

S-27

The second way to solve this is to use the method of undetermined coefficients from beginning to end. When the forcing function is x times sine or cosine of x (and D2 + 1 is not a factor of the operator), then the particular integral will be in the form of p (x) = λc0 cos x + λc1 x cos x + λs0 sin x + λs1 x sin x.

(a)

By applying Leibniz’s rule we find that d3 p = λc0 sin x + λc1 x sin x − 3λc1 cos x − λs0 cos x − λs1 x cos x − 3λs1 sin x. dx3 3

d p When you take the difference, dx 3 − p = x sin x, then pick the result apart into separate equations according to whether the coefficient is multiplied by cos x or x cos x or sin x or x sin x, it gives the linear system, λc1 − λs1 = 1, − λc1 − λs1 = 0, − λc0 − 3λc1 − λs0 = 0, λc0 − λs0 − 3λs1 = 0.

Solving the above yields λc0 = − 32 , λc1 = coefficients in (a) gives the book’s answer.

1 2,

λs0 = 0, and λs1 = − 21 . Using these

Page 315 (§ 130. Particular solutions) case 6, example 2: If you tried the method given on the you probably ran into difficulty. This
top half of page 315 directly,
is because D2 + 1 is a factor of D4 − 1 , making sin x a part of the complementary solution. A way to do this is to solve it in two phases.
−1
−1
−1 D4 − 1 x sin x = D2 + 1 D2 − 1 x sin x.
Applying (11) on page 315 to the D2 − 1 factor, o
−1
−1
−1
−1 n
−1 x − D2 − 1 (2D) D2 − 1 D2 + 1 D2 − 1 x sin x = D2 + 1 sin x. By (9) on page 313 this is o
−1
−1
−1
−1 n D2 + 1 D2 − 1 x sin x = − 21 D2 + 1 x − D2 − 1 (2D) sin x, o
−1
−1 n = − 21 D2 + 1 cos x . x sin x − 2 D2 − 1 By (10) on page 313
−1
−1
−1 {x sin x + cos x} . D2 + 1 D2 − 1 x sin x = − 12 D2 + 1

By superposition n o
−1
−1
−1 x sin x − D2 + 1 D2 − 1 x sin x = − 12 D2 + 1

1 2

n

D2 + 1


−1

o cos x .

By the development on page 314 n o
−1
−1
−1 D2 + 1 D2 − 1 x sin x = − 21 D2 + 1 (a) x sin x − 14 x sin x.
−1 For the second phase, it remains to find D2 + 1 x sin x. Unfortunately this is the exceptional case of (11) on page 315 that does not immediately yield to this method. You
−1 end up with an expression in which you must, once again, evaluate D2 + 1 x sin x. To continue you need to apply a trick. Here then is that trick. n o
−1
−1
−1 D2 + 1 x sin x = x − D2 + 1 (2D) D2 + 1 sin x, n o
−1
= x − D2 + 1 (2D) − 21 x cos x,
−1 D (x cos x) , = − 21 x2 cos x + D2 + 1
−1 2 1 2 = − 2 x cos x + D + 1 (−x sin x + cos x) .

S-28

Now add D2 + 1


−1

x sin x to both sides of the above:
−1
−1 2 D2 + 1 x sin x = − 21 x2 cos x + D2 + 1 cos x, = − 21 x2 cos x + 21 x sin x,

from which we see that D2 + 1


−1

x sin x = − 41 x2 cos x + 14 x sin x.

We can also use the method of undetermined coefficients to solve this. In this case, rather than set it up formally, I will use the inspection method (that is making educated
−1 guesses and refining them). Because − 21 x cos x results when you apply D2 + 1 to sin x, we should suspect that when we apply the same to x sin x, one of the terms of the result ought to be a multiple of x2 cos x. So we apply the operator to that expression.
   2  D2 + 1 x2 cos x =  −x cos x − 4x sin x + 2 cos x +  x2 cos x.

We need to add another term that will cancel the 2 cos x term in the above. Based upon experience, we suspect that term will be a multiple of x sin x.

D2 + 1 x2 cos x + λx sin x =      2    −x cos x − 4x sin x + 2 cos x −  x2 cos x + λx sin x + 2λ cos x +  λx sin x. 

Clearly if λ = −1, then D2 + 1 x2 cos x − x sin x = −4x sin x. From this we conclude
−1 again that D2 + 1 x sin x = − 41 x2 cos x + 41 x sin x. We substitute this into (a) on the previous page to get
−1
−1 D2 + 1 D2 − 1 x sin x = 18 x2 cos x − 81 x sin x − 41 x sin x. Gathering terms in the above yields the book’s answer.

Page 315 (§ 130. Particular solutions) case 6, example 3: To solve this

one it is useful first to evaluate D2 − 1 ex sin x and D2 − 1 ex cos x. This is done easily using the method of undetermined coefficients. In each case we assume that p (x) = λc ex cos x + λs ex sin x. We find
D2 − 1 p (x) = (2λs − λc ) ex cos x + (−2λc − λs ) ex sin x. Separating terms according multiples of ex cos x and ex sin x and putting it all into matrix form we have λc λs    
−1 x −1 2 1 row for ex cos x for finding D2 − 1 e cos x. x −2 −1 0 row for e sin x By Cramer’s rule, λc = − 51 and λs = 25 . Hence
−1 x D2 − 1 e cos x = − 51 ex cos x + 25 ex sin x.

(a)

Also

λc 

λs

−1 2 −2 −1



0 1



row for ex cos x row for ex sin x



for finding D2 − 1


−1

ex sin x.

By Cramer’s rule, λc = − 52 and λs = − 51 . Hence
−1 x D2 − 1 e sin x = − 52 ex cos x − 15 ex sin x. (b)
2 x The problem is to solve D − 1 y = xe sin x. Clearly the complementary solution is yc = C1 ex + C2 e−x . To find the particular solution, p (x), we apply (11) on page 315. n o
−1 x
−1
−1 x p (x) = D2 − 1 e sin x = x − D2 − 1 (2D) D2 − 1 e sin x.

S-29

By (b) above we have n o

−1 p (x) = x − D2 − 1 (2D) − 52 ex cos x − 15 ex sin x
−1
= − 52 xex cos x − 51 xex sin x − 2 D2 − 1 − 35 ex cos x + 51 ex sin x .

Finally by superposition and by (a) and (b) above we have p (x) = − 52 xex cos x − 51 xex sin x −

2 x 25 e

cos x +

14 x 25 e sin x.

1 x e and gathering terms from the above yields the book’s answer. Factoring out − 25

Page 315 (§ 130. Particular solutions) case 6, example 4: To solve this one
−1
−1 it is useful first to evaluate D2 − 1 x cos x and D2 − 1 x sin x. Each of them succumb easily to combining (11) on page 315 with (9) and (10) on page 313. n o
−1
−1
−1 D2 − 1 x cos x = x − D2 − 1 (2D) D2 − 1 cos x, n o
−1
= x − D2 − 1 (2D) − 12 cos x ,
−1 = − 12 x cos x + D2 − 1 D cos x,
−1 = − 12 x cos x − D2 − 1 sin x, 1 2

= − 12 x cos x +

sin x.

(a)

Likewise D2 − 1


−1

n o
−1
−1 x sin x = x − D2 − 1 (2D) D2 − 1 sin x, o n

−1 (2D) − 21 sin x , = x − D2 − 1
−1 = − 12 x sin x + D2 − 1 D sin x,
−1 2 1 cos x, = − 2 x sin x + D − 1 = − 12 x sin x −

1 2

cos x.

(b)

Now apply (11) on page 315 then (a) above and finally (b) above to the problem at hand. n o
−1 2
−1
−1 D2 − 1 x cos x = x − D2 − 1 (2D) D2 − 1 x cos x, n o

−1 = x − D2 − 1 (2D) − 12 x cos x + 12 sin x ,
−1 = − 12 x2 cos x + 21 x sin x + D2 − 1 D (x cos x − sin x) ,
−1 x sin x, = − 12 x2 cos x + 21 x sin x − D2 − 1
1 2 1 1 = − 2 x cos x + 2 x sin x − − 2 x sin x − 21 cos x , = − 12 x2 cos x + x sin x +

which is the book’s answer.

S-30

1 2

cos x,

The Method of Laplace. Section § 130 details a variety of artifices you can employ to determine the particular integral when you have a linear differential equation with constant coefficients together with a forcing function. I have also described, here in the supplement, the additional method of undetermined coefficients, which Mellor does not cover, as well as stating the principle of superposition. By applying these methods, you can find the particular integral for any such equation whenever the forcing function is any sum and/or product of exponentials, sines, cosines, and polynomials. The only difficulty is that the method you need to bring to bear depends on the type of forcing function. Here I will describe the method of Laplace, with which you can find the particular integral in any problem using only this single method. The method of Laplace has the additional advantage in that it can deal with initial conditions and it can easily be applied to simultaneous equations. It is, to quote Tolkein, the one ring to rule them all. You may have noticed that there is a correspondence between functions of the equation’s independent variable (usually x or t), and the functions that result from the differential operator, D. For example, if you have 1 dy − ay = ebx hence y = ebx , dx D−a You know that the complementary function will have an eax term in it. If we were to find the particular integral by integrating, the integral expression to be solved would be Z bx e p (x) = eax dx. eax We see that there is a correspondence between seeing the (D − a)−1 as an operator and the function, eax . This correspondence works in both directions. So we can observe that since the forcing function is ebx in the example, we should be able to replace it with −1 (D − b) . We end up solving for y as a function of D instead of as a function of x.   1 1 1 1 . = − Y (D) = (D − a) (D − b) a−b D−a D−b Notice that when you go from a function of x (or t) to a function of D, as we did with y, it is customary notation to use the upper case symbol for the function of D (e.g., Y (D)). Once the D-function has been expanded into partial fractions, it is immediately evident what the final solution to this problem is. We simply translate from the D-function −1 back to a function of x. We know that (D − a) corresponds with eax , which is a part of the complementary function. So that part of the D-function becomes Ceax . The −1 (D − b) portion translates back to ebx . Since it is a part of the particular integral, we must pay attention to its constant multiplier and carry that into the final solution. Hence the solution to the example is 1 bx y = Ceax − e . a−b The table on the following page shows the pairings of various functions of x with their corresponding functions of D. It also shows some rules by which you can infer the corresponding D-function of more complicated functions of x. If you have a function of x, then its corresponding D-function is known as its Laplace transform. Many if not most modern textbooks use the symbol, s (or sometimes p), in place of D, as the independent variable of a function’s Laplace transform. This is a matter of nomenclature only, and it does not change the essence of the method one bit. Because Mellor uses the symbol, D, throughout, I will stick with it here in the further explanation of the Laplace method. What the Laplace method does is to convert the problem of finding a particular integral into nothing more than a partial fractions problem. Take, for example, the problem
d2 y + 9y = x2 , or equivalently, D2 + 9 y = x2 . 2 dx (text continues on page S-33).

S-31

f (x)

F (D)

1

1 D 1 D−a

eax sin bx

b D 2 + b2

cos bx

D D 2 + b2

xn , where n ≥ 0

n! Dn+1 n!

xn eax , where n ≥ 0 Rule:

eax g (x)

Hence

eax sin bx

n+1

(D − a)

G (D − a) b (D − a)2 + b2 D−a

and eax cos bx

2

(D − a) + b2

xn g (x)

Rule:

xeax sin bx

Hence

n

and xeax cos bx

Rule:

n

g (ax) dg dx

Rule: Z

dn G dDn

n

(−1)

2b (D − a) 2

(D − a) + b2 2

o2

(D − a) − b2 o2 2 (D − a) + b2 1 G a



D a



(D) G (D)

x

0

1 G (D) D

Rule: µg (x) + νh (x)

µG (D) + νH (D)

Rule:

g (u) du

sinh bx cosh bx

D2

b − b2

D2

D − b2

Short table of Laplace Transforms.

S-32

Using the table to look up the forcing function and solving for Y (D), it becomes 2D 2 2 2 = . − + Y (D) = (D2 + 9) D3 81 (D2 + 9) 81D 9D3 We know that the complementary function for the original equation is C1 sin 3x + C2 cos 3x. Once expanded into partial fractions, the first term of Y (D) translates back to a multiple of cos 3x, so it is a part of the complementary function. The remaining terms are a part of the particular integral. Using the table to translate them back to a function of x, we find the full solution is y = C1 sin 3x + C2 cos 3x −

2 81

+ 19 x2 .

A type of example that tends to trip up beginners is where the forcing function is linearly dependent upon a term of the complementary function. The Laplace method handles these seamlessly. For example
d2 y + b2 y = cos bx, or equivalently, D2 + b2 y = cos bx. 2 dx The complementary function is C1 sin bx + C2 cos bx, so clearly the forcing function is linearly dependent upon a term of the complementary function. Using the table to solve this one for Y (D), D Y (D) = 2. 2 (D + b2 ) Using the table to translate back we see that the particular integral in this case is 1 x sin bx, which is consistent with the result on page 314 of the text (note that p (x) = 2b the table entry for xeax sin bx gives x sin bx when a = 0). If we had started with a forcing function of sin bx, which is also linearly dependent upon the complementary function, the solution for Y (D) would have been b Y (D) = 2, 2 (D + b2 ) b2 1 = , b (D2 + b2 )2 1 21 b2 − 12 D2 + 21 D2 + 12 b2 , 2 b (D2 + b2 ) ( ) D 2 − b2 1 1 . − 2 =− 2b (D2 + b2 )2 D + b2 =

The second term inside the brackets is a part of the complementary function. The first 1 term corresponds in the table to x cos bx, so the particular solution is p (x) = − 2b x cos bx, which is consistent with example 1 on page 314. As stated before, the method of Laplace can also determine the undetermined constants in the complementary function given initial conditions at x (or t) equal to zero. If y is a function of x (or t), then the Laplace transform of the first derivative of y is DY (D) − y (0). The Laplace transform of the second derivative of y is D2 Y (D) − Dy (0) − y 0 (0), where y (0) and y 0 (0) are initial conditions at x = 0 (or t = 0). Higher derivatives follow the same pattern. As an example, we solve the following equation with a stipultated initial condition: dy + ay = sin ωt, where y (0) = 1. dt In the Laplace transform version, this equation is ω . (D + a) Y (D) − 1 = D2 + ω 2 Make sure you understand how the initial condition results in the −1 just to the left of the equal.

S-33

Solving for Y , ω

1 , + + a) D+a   1 a−D 1 ω + + . (a) = 2 2 2 2 a +ω D +ω D+a D+a ω Letting A = a2 +ω 2 , we find that a y (t) = A sin ωt − A cos ωt + (A + 1) e−at . ω The solution above derives from looking up the terms of (a) on the table. As an exercise, you are encouraged to confirm that y (0) = 1 and to verify that this solution satisfies the original equation. Observe that there are no undetermined constants in this solution. This is because we stipulated the initial condition, and doing so fixes the multiplier of the complementary function. Y (D) =

(D2

+

ω 2 ) (D

Applying the Laplace method to simultaneous equations. Here is an example. dy + z = 0, where y (0) = 0, dx dz − 4y = x, where z (0) = −3. dx Translating this into Laplace form, DY

+

Z

=

DZ

−

4Y

=

0, 1 − 3. D2

In matrix form this is Y

Z

D

1

−4

D

!

0

!

1 −3 D2 The determinant of this matrix is D2 + 4. Cramer’s rule yields   3 1 1 1 3 1 Y (D) = , − 2 − 2 2 = − D2 + 4 D (D + 4) D2 + 4 4 D2 D +4   3D 3D 1 1 D 1 − 2 − 2 = − 2 . Z (D) = D (D2 + 4) D +4 4 D D +4 D +4 Gathering terms and then translating back using the table gives y= z=

13 1 8 sin 2x − 4 x, 13 1 4 − 4 cos 2x.

Again it is left as an exercise for you to verify that this solution satisfies both the original system of equations and the initial conditions. In engineering, the problem of simultaneous linear equations occurs over and over again in analog electronics and in feedback and control theory. In those fields, Laplace is the preferred method for solving such problems. Indeed manufacturers of analog electronic parts and electro-mechanical transducers frequently characterize their products by citing product responses in terms of Laplace functions. The power of the Laplace method is that it converts calculus problems into algebra problems. You are urged to review the example above to see how solving for Y and Z became, in the Laplace domain, is simply a problem of solving a system of linear algebra equations. Decomposing that solution into partial fractions is again just an algebra problem. The final step of translating back is nothing more than a matter of looking it up on the table.

S-34

Page 315 (Linear equations with variable coefficients) case 1, info: The funnylooking symbol, ϑ, that the text uses throughout this case is called vartheta, and is a cursive form of the Greek letter, θ. Page 316 (§ 131. Linear equations with variable coefficients) case 1, first example 1: Page 315 establishes that z = log x. Hence dz 1 1 d dz d d = , so by the chain rule, = = . dx x x dz dx dz dx Multiplying through by x completes the proof. Page 316 (§ 131. Linear equations with variable coefficients) case 1, first example 2: The ϑ operation consists of first taking the derivative, then multiplying by x. d d (xm ) = mxm−1 ; Now multiplying by x, x (xm ) = mxm . dx dx Alternatively, from page 315, x = ez , so xm = emz . By example 1, d mz ϑ (xm ) = (e ) = memz = mxm . dz Page 316 (§ 131. Linear equations with variable coefficients) case 1, second example 2: Using the ϑ operator according to identities at the bottom of page 315,
ϑ ϑ−1 ϑy  + q 2 y = 0, ∴ ϑ2 y + q 2 y = 0.
y + Per page 315, we make the substitution, z = log x, and apply the identity from ex
d , you have D2 + q 2 y = 0. Hence ample 1. So with the substitution and D = dz y = C1 sin qz + C2 cos qz = C1 sin (q log x) + C2 cos (q log x), which is the book’s answer. Page 316 (Linear equations with variable coefficients) case 1, third example 2: Substituting xm for y, m (m − 1) xm + 4mxm + 2xm = 0.

Dividing out xm and solving for m:

m2 + 3m + 2 = 0, −1

The solution, then, is y = C1 x

hence −2

+ C2 x

m = −1 or m = −2.

, which is is the book’s answer.

Observe that we could also have applied this method to the second example 2 (that is the previous example shown on this page), but m would have been imaginary in that case. For that example we would have gotten m = ±ιq, for a solution of y = A1 xιq + A2 x−ιq . As an exercise, apply Euler’s formula to show that this solution is equivalent to the solution already shown of y = C1 sin (q log x) + C2 cos (q log x). What is the relationship between the undetermined constant sets, A1 , A2 and C1 , C2 ? Page 316 (§ 131. Linear equations with variable coefficients) case 1, example 3: Replacing x with ez , we have: {D (D − 1) − 3D + 4} y = e3z , 2

(D − 2) y = e3z . 2z

2z

(a) z

The complementary function is y = C1 e + C2 ze . Replacing e with x and z with log x yields y = C1 x2 + C2 x2 log x, which is equivalent to the book’s answer. In addition, we can apply the method of undetermined coefficients to (a) to establish the particular integral. Let p (x) = λe3z . Then
2 (D − 2) p (x) = D2 − 4D + 4 p (x) = (9 − 12 + 4) λe3x = e3x ,

hence λ = 1. So p (x) = e3x = x3 .

S-35

Page 316 (§ 131. Linear equations with variable coefficients) case 1, example 4: The auxillary equation for xm is m2 − m − 2 = 0. So m = 2 or m = −1. That leads to y = C1 x2 + C2 x−1 , which is the book’s answer. Page 317 (§ 131. Linear equations with variable coefficients) case 1, example 1: The equation is equivalent to {ϑ (ϑ − 1) − 2ϑ − 4} y = x4 . This factors into (ϑ + 1) (ϑ − 4) y = x4 .

The roots are 4 and −1, which establishes the complementary solution as C1 x4 + C2 x−1 . Substituting x = ez and ϑ with D, we can find the particular integral by solving −1

p (z) = (D + 1)

−1 4z

(D − 4)

e .

According to (5) on page 312, this is the same as

−1

p (z) = e4z (D + 5) Applying the D−1 first,

D−1 · 1. −1

p (z) = e4z (D + 5) z.
1 From this you get p (z) = e4z 51 z − 25 . But e4z = x4 is a part of the complementary 1 4z solution. So we can drop the 25 e term. Substituting z = log x gives p (x) = 15 x4 log x. Page 317 (§ 131. Linear equations with variable coefficients) case 1, example 4: This one is equivalent to {ϑ (ϑ − 1) + 4ϑ + 2} y = ex ,
ϑ2 + 3ϑ + 2 y = ex , (ϑ + 1) (ϑ + 2) y = ex . Hence the complementary function is C1 x−1 + C2 x−2 . The book offers no direct method for finding the particular integral of this one. But we can cobble together a procedure out of various methods we’ve learned so far. First we replace x with ez and ϑ with D, as we have done in other examples. z −1 −1 p (z) = (D + 2) (D + 1) e(e ) . −1

Then we apply integral in (4) from page 311 to the (D + 1) operator. Z z z −1 −z −1 p (z) = (D + 2) e ez e(e ) dz = (D + 2) e−z e(e ) . −1

Now do the same with the (D + 2)

−1 −z (ez )

p (z) = (D + 2)

e

e

operator. Z z z −2z = e e2z e−z e(e ) dz = e−2z e(e ) .

Back-substitute x for ez , and you get the book’s answer.

Taking a more general view of the procedure we used on this example, we solve p (x) = (ϑ + a)−1 R (x) by solving p (z) = (D + a) R (ez ). We do this by applying the integral in (4) on page 311. Z −az p (z) = e eaz R (ez ) dz.

Since log x = z, we also have dx/x = dz. Back-substituting this integral gives the Z formula, −1 −a p (x) = (ϑ + a) R (x) = x xa−1 R (x) dx, (a) −1

which is a general formula for such problems. If you apply (ϑ + 1) to R (x) = ex −1 using (a), then apply (ϑ + 2) to the result, you will end up with the same particular integral for example 4 as we arrived at above.

S-36

−1

Suppose we do the same example, but this time we first apply (ϑ + 2) to R (x) = ex , −1 then apply (ϑ + 1) to the result. We expect to end up with the same function as Z before. ex ex −1 − 2. (ϑ + 2) ex = x−2 x ex dx = x−2 (x − 1) ex = x x Before we apply the second operator to this result, we must deal with the fact that we will end up with integrals that cannot be expressed in closed form. Let Z x Z x e e E1 (x) = dx. dx and E2 (x) = x x2 Using integration by parts (letting u = ex , du = ex dx, dv = dx/x2 , and v = −1/x), Z Z Z x ex e ex E2 (x) = u dv = uv − v du = − + dx = E1 (x) − . x x x −1

We now apply (ϑ + 1) to ex /x − ex /x2 , using (a) from the previous page and the above identity to substitute for E2 (x).   x     x Z x e e ex ex −1 − + e E1 (x) E1 (x) , (ϑ + 1) − 2 = x−1 x0 − 2 dx = x−1  x x x x x demonstrating that we get the same result no matter in which order we apply the operators. The more common class of forcing function for problems like this is a polynomial. If we apply (a) from the previous page to R = xm , we get Z xm −1 p (x) = (ϑ + a) xm = x−a xa+m−1 dx = for a + m 6= 0. (b) a+m When a + m = 0, you have p (x) = x−a log x. The formula (b) also works for negative and noninteger exponents. This formula is applicable to the remaining examples in this case (5 through 8). Page 317 (§ 131. Linear equations with variable coefficients) case 1, example 5: Equation becomes {ϑ(ϑ − 1)(ϑ − 2) + 2ϑ(ϑ − 1) − ϑ + 1} y = x + x3 ,  3  − ϑ + 1 y = x + x3 ,  + 2ϑ2 −  2ϑ 2ϑ ϑ − 3ϑ2 +   3 ϑ − ϑ2 − ϑ + 1 y = x + x3 , (ϑ − 1)2 (ϑ + 1) y = x + x3 . n

Observe that in general, a repeated root, (ϑ − r) , produces terms in the complementary k function of Ck xr (log x) for k = 0 to k = n − 1. So in this example, the complementary function is C0 x + C1 x log x + C2 x−1 , which is equivalent to the book’s complementary function. To find the particular integral, we apply (b) on this page, and where necessary, (a) on the previous page to
−2 −1 p (x) = (ϑ − 1) (ϑ + 1) x + x3 ,   x x3 −2 = (ϑ − 1) by (b), + 2 4   x x3 −1 = (ϑ − 1) log x + by (b), 2 8 Z log x x3 x dx + by (a) on the first term and (b) on the second, = 2 x 16 x x3 2 = (log x) + , which is the book’s answer. 4 16

S-37

Page 317 (§ 131. Linear equations with variable coefficients) case 1, example 6: Equation becomes 10 {ϑ(ϑ − 1)(ϑ − 2) + 2ϑ(ϑ − 1) + 2} y = 10x + , x  3 10 2 2  + 2 y = 10x + ,  + 2ϑ −  2ϑ 2ϑ ϑ − 3ϑ +  x  3 10 2 ϑ − ϑ + 2 y = 10x + , x
10 2 (ϑ + 1) ϑ − 2ϑ + 2 y = 10x + . x The second factor above has no real roots. Its nonreal roots are ϑ = 1 ± ι. So the complementary function is C1 x−1 + C2 x1+ι + C3 x1−ι . To turn the second and third terms into real form, C2 x1+ι + C3 x1−ι = C2 xeι log x + C3 xe−ι log x , and then by Euler’s formula = C2 x {cos (log x) + ι sin (log x)} + C3 x {cos (log x) − ι sin (log x)} .

If you stipulate that C2 and C3 be complex complements of each other, then the above is equal to C20 x cos (log x) + C30 x sin (log x), where both C20 and C30 are both real. Putting the pieces together we find that we now have the same complementary function as the book. To find the particular integral, the setup is   10 . p (x) = (ϑ + 1)−1 (ϑ − 1 − ι)−1 (ϑ − 1 + ι)−1 10x + x

Formula (b) on the previous page works even when the ϑ-factor is nonreal. Applying the second and third factor to each of the forcing function terms using (b) yields  
10x 10x−1 −1 −1 p (x) = (ϑ + 1) = (ϑ + 1) 10x + 2x−1 . + (−ι) (ι) (−2 − ι) (−2 + ι) −1

Finally using (b) to apply (ϑ + 1)

−1

p (x) = (ϑ + 1)

to each of the terms that result,
10x + 2x−1 = 5x + 2x−1 log x.

Putting this together with the complementary function yields the book’s answer. Page 317 (§ 131. Linear equations with variable coefficients)
case 1, example 8: The third example 2 on page 316 uses the operator, ϑ2 + 3ϑ + 2 = (ϑ + 1) (ϑ + 2). So the setup for finding the particular integral when R = x−1 is −1

p (x) = (ϑ + 1)

−1

(ϑ + 2)

x−1 ,

and by (b),

−1

x −1 = (ϑ + 1) x−1 . by (b) again, 2−1 = x−1 log x, which is the book’s answer. −1

= (ϑ + 1)

Page 317 (Linear equations with variable coefficients) case 2, example 1: A better substitution than the one suggested in the book for this is z = a/b + x. This is because all derivatives of y with respect to z are equal to the corresponding derivatives with respect to x. Equation becomes
 2 dy d2 y b2 z 2 2 + b2 z + c2 y = 0 ≡ b ϑ (ϑ − 1) + b2 ϑ + c2 y = b2 ϑ2 − c2 y = 0. dz dz By the methods given for case 1, the solution to this is C10 sin (c/b log z)+C20 cos (c/b log z). See page 316, second example 2, which is worked on page S-35. Back-substitute for z to obtain at C10 sin (c/b log (a/b + x)) + C20 cos (c/b log (a/b + x)), which is the same as  c  c c c log (a + bx) − log b + C20 cos log (a + bx) − log b . C10 sin b b b b Use trig identities for sin (u − v) and for cos (u − v) to arrive at the book’s answer.

S-38

Page 317 (§ 131. Linear equations with variable coefficients) case 2, example 2: Substitute z = x + a. Equation becomes d2 y dy − 4z + 6y = x ≡ {ϑ (ϑ − 1) − 4ϑ + 6} y = (ϑ − 2) (ϑ − 3) y = z − a. dz 2 dz 2 3 Clearly the complementary function for this is C1 z 2 + C2 z 3 = C1 (x + a) + C2 (x + a) . The setup for finding the particular integral is z2

−1

p (z) = (ϑ − 2)

−1

(ϑ − 3) (z − a) , which by (b) on S-37 is    z a z a = (ϑ − 2)−1 − + ; by (b) again, = (ϑ − 2)−1 − (−3 + 1) (−3 + 0) 2 3 a z a x+a a x a z + = − = − = + , = − 2 (−2 + 1) 3 (−2 + 0) 2 6 2 6 2 3 which, when combined with the complementary function, is the book’s answer. Page 318 (§ 132. Exact equations) clarification: You might have found book’s development of how to deal with higher order exact equations to be confusing. The easy way to understand it is to work it backwards from the first integral. If you begin with R d2 y dy X0 2 + (X1 − X00 ) + (X2 − X10 + X000 ) y = R dx + C, dx dx then take its derivative using the product rule, you get X0

2 d3 y 0 d y + X + 0 dx3 dx2 d2 y dy (X1 − X00 ) 2 + (X10 − X000 ) + dx dx dy + (X20 − X100 + X0000 ) y = R, (X2 − X10 + X000 ) dx

d2 y dy d3 y + X + X2 + X3 y = R, 1 3 2 dx dx dx which holds provided that X3 = X20 − X100 + X0000 . So you can see that by differentiating (5) on page 318, we have recovered (1) on page 317. At the same time we have also evolved the exactness criterion shown in (4) of page 318. This method can easily be extended to determine exactness and first integrals of equations of order higher than 3. X0

Page 318 (§ 132. Exact equations) example 2: We have X3 = 2, X20 = 4, X100 = 2, and X0000 = 0. Hence X3 − X20 + X100 − X0000 = 2 − 4 + 2 − 0 = 0, so the equation is exact. By (5) on page 318, the first integral is
dy d2 y x 2 + x2 − 3 − 1 + (4x − 2x) y = C1 . dx dx If you allow the coefficient functions of the above to be Y0 = x, Y1 = x2 −4, and Y2 = 2x, then Y2 − Y10 + Y000 = 2x − 2x = 0. So this new equation is also exact. We apply (5) again to get  
dy dy 5 C2 2 x + x − 4 − 1 y = C1 x + C2 ≡ + x− . y = C1 + dx dx x x Using the method on page 294, the solution to the above is   Z
2 2 C1 x−5 + C2 x−6 ex /2 dx . y = x5 e−x /2 C3 +

The integral in the expression above cannot be expressed in elementary functions, but can be described using something called the incomplete gamma function. You can find more about incomplete gamma in Handbook of Mathematical Functions by Abramowitz and Stegun (U.S. Department of Commerce, National Bureau of Standards, Applied Mathematics Series · 55).

S-39

Page 318 (§ 132. Exact equations) example 3: Example 1 has already demonstrated that this equation is exact. So we apply (5) on page 318 to find the first integral. 2 d2 y dy 4 dy 3 5d y + (15 − 5) x + 10x4 + (60 − 60 + 20) x y = x + 20x3 y = ex + C. dx2 dx dx2 dx If we let Y0 = x5 , Y1 = 15x4 , and Y2 = 20x3 , then Y2 = Y10 − Y000 , hence this equation is also exact. So we apply (5) again to get the second integral. dy dy + (10 − 5) x4 y = x5 + 5x4 y = ex + Cx + C2 . x5 dx dx This equation is also exact. Its left side is a perfect differential.
d x5 y = ex + Cx + C2 ∴ x5 y = ex + C1 x2 + C2 x + C3 , dx with C1 = 12 C, which is the book’s answer.

x5

Observe that if you divide the original equation by x2 , the resulting equation is homogeneous in powers, d3 y d2 y dy ex x3 3 + 15x2 2 + 60x + 60y = 2 , dx dx dx x and can be solved using the method from § 131. ex {ϑ (ϑ − 1) (ϑ − 2) + 15ϑ (ϑ − 1) + 60ϑ + 60} y = 2 , x
ex 3 2 ϑ + 12ϑ + 47ϑ + 60 y = 2 , x ex (ϑ + 5) (ϑ + 4) (ϑ + 3) y = 2 . x The complementary function is C1 x−3 + C2 x−4 + C3 x−5 . To find the particular integral, we need to solve the following by repeated application of (a) on page S-36: ex (ϑ + 5) (ϑ + 4) (ϑ + 3) p (x) = 2 , x Z ex ex (ϑ + 5) (ϑ + 4) p (x) = x−3 x2 2 dx = 3 , x x Z x ex e (ϑ + 5) p (x) = x−4 x3 3 dx = 4 , x x Z x e ex p (x) = x−5 x4 4 dx = 5 . x x Combining this result with the complementary function, we get a solution for y that is equivalent to the book’s answer. Page 319 (§ 132 Exact equations) Comments on Forsyth’s elimination of terms: That xm dn y/dxn is a perfect differential whenever m < n is proved by taking the integral of this expression by parts: Z Z n−1 n−1 n y y md m−1 d md y dx = x − m x dx. x dxn dxn−1 dxn−1 Note that to the right of the equal both the exponent and the order of the derivative of the expression under the integral have been reduced by unity. So by repeating the process over and over, we can see that the exponent reaches zero before the order of the derivative does, provided m < n. At that point only one more integration is required, and since we would be integrating only a derivative of y (not multiplied by any polynomial of x) we would simply get the next lower derivative (or y itself if m = n − 1). An easy way to justify eliminating such terms from the exactness criterion, (4) on page 318, is that the order of the derivative we’d have to take of Xk would exceed its highest exponenent of x, and so that derivative would be zero and contribute nothing to the exactness criterion formula.

S-40

Page 320 (§ 133. Equations with missing terms) example 1: Back-substituting p with dy/dx into the text’s result, p2 y 2 = y 4 + C 4 ,  2 p dy dy 2 y = y 4 + C 4 hence y = y4 + C 4 . dx dx Now let u = y 2 , and 12 du = y dy, the above becomes 1 du p 2 du = 2dx. = u + C 4 , which by separation of variables is √ 2 2 dx u + C4 Integrating,  u  u sinh−1 = 2x + C2 hence = sinh (2x + C2 ) . 2 C C2 Back-substituting y 2 for u, multiplying by C, and changing the symbol, C, to C1 results in the book’s answer.

Page 320 (§ 133. Equations with missing terms) example 2: This equation is linear with constant coefficients and constant forcing function, and as such it can be solved by methods developed in § 129 and § 130. Doing it using the method suggested in the example, we divide through by µ and make the substitution, observing that d2 x1 /dt2 = d2 x/dt2 . Equation becomes 1 d2 x1 + x1 = 0 hence µ dt2

d2 x1 + µx1 = 0. dt2 √
√
You can solve this by inspection: x1 = C1 cos t µ + C2 sin t µ . Back-substitute x1 = x − ν/µ to arrive at the book’s answer.

Page 320 (§ 133. Equations with missing terms) example 4: Let p = dy/dx. Equation becomes dp dp = a dx. − ap2 = 0, which by separation of variables is dx p2 Integrating then back-substituting, dy 1 1 =− . − = ax + C hence p dx ax + C Integrating again gives y = − (1/a) log (ax + C) + C 0 . Multiply both sides by −a, then take the exponential function of both sides to get, 0

e−ay = e−aC (ax + C) = C1 x + C2

where C1 = ae−aC

0

0

and C2 = e−aC C,

which is the book’s answer. Page 320 (§ 133. Equations with missing terms) example 5: Substitute p for dy/dx and p dp/dy for d2 y/dx2 , and the equation becomes dp dy p dp 1 + p2 = yp , which by separating variables is = . dy y 1 + p2 Integrating, p
hence C1 y = 1 + p2 where C1 = eC . log y + C = 21 log 1 + p2

Back-substitute dy/dx for p, s C1 y =

1+



dy dx

2

hence

q dy C12 y 2 − 1 = . dx

Separating variables and integrating, Z Z dy cosh−1 (C1 y) p = x + C2 . = dx hence C1 C12 y 2 − 1 Letting a = C1 and b = C1 C2 , and multiplying by a, then taking cosh of both sides, then dividing both sides by a, the above results in the book’s answer.

S-41

Page 320 (§ 133. Equations with missing terms) example 6: The book’s answer is evident by inspection using methods from earlier sections. But doing it using the method from this section, we substitute p for dV /dx and p dp/dV for d2 V /dx2 . Equation becomes dp p − β 2 V = 0 or equivalently p dp = β 2 V dV. dV Integrating and cancelling the common factor of 21 , p C p2 = β 2 V 2 + C hence p = β V 2 + C 0 where C 0 = . β Back-substituting and separating variables, p dV dV = β dx. = β V 2 + C 0 hence √ dx V 2 + C0 Integrating,   √ V −1 √ = βx + C 00 hence V = C 0 sinh (βx + C 00 ) . sinh 0 C Using the exponential identity to expand sinh,  √  00 00 V = 12 C 0 eβx+C − e−βx−C . √ √ 00 00 Let C1 = 12 C 0 eC and C2 = − 21 C 0 e−C to obtain the book’s answer. The question 0 00 then arises, can you choose C and C in such a way as to arrive at an arbitrary C1 and 00 00 C2 ? This would be true provided that the ratio of eC to e−C can be made to be any desired value, positive or negative, by proper choice of C 00 . So if for arbitrary A, 00

log A eC 2C 00 = A, then C 00 = . 00 = e −C e 2 If we allow nonreal values for C 00 (such as C 00 = 21 log 2 + 12 ιπ for A = −2 for example) then by Euler’s formula, we can solve this given any A. Also by allowing√ C 0 to be imaginary when desired, we can satisfy the cases where we would want C 0 to be negative. Page 322 (§ 133. Equations with missing terms) example 5: Letting p = dy/dx, the equation becomes Z Z dp dx x = 1, or equivalently dp = . dx x Integrating yields p = log x + C. Back-subsituting for p, Z Z dy = log x + C, or equivalently dy = log x + C dx. dx By integrating the above and letting C1 = C − 1 you get the book’s answer. Page 335 (§ 135. Chemical Reaction Rates) solving (16): The equation to be solved is a−z dy z−y dy y z dz = , or equivalently = or + = . dy z−y dz a−z dz a−z a−z The last version is in the form given by (1) of § 122 (page 296). Hence the integrating 1 . The exact equation is factor for this is µ = e− log(a−z) = a−z dy y−z + 2 dz = 0. a−z (a − z) Integrating the second term dz yields the general solution, y a − − log (a − z) = C. a−z a−z Substituting the conditions on page 335, a = 1, y = 0, and z = 0, into the above forces C = −1. The solution, (17) on page 335, follows from that.

S-42

Page 335 (§ 135. Chemical Reaction Rates) solving (18): The equation to be solved is dz a−z dx y−x dy x y = , or equivalently = or + = , dx y−x dz a−z dz a−z a−z where y = z + (a − z) log (a − z). Like the last equation, this is first-order linear, with 1 . Multiplying by the integrating factor the integrating factor being µ = e− log(a−z) = a−z and substituting for y, z + (a − z) log (a − z) dx x dz dz, + 2 = 2 a−z (a − z) (a − z) ) ( log (a − z) z + = dz, a−z (a − z)2 which is exact. Integrating yields x a 2 = log (a − z) + − 1 {log (a − z)} + C. a−z a−z 2 Using the analogous initial conditions as were stipulated for (16), that is a = 1, x = 0, and z = 0, we find again that C = −1. Substitute −1 for C, then multiply the above through by a − z and take the cancellation to arrive at the book’s answer. Page 336 (§ 135. Chemical Reaction Rates) example: (14) and (15) on page 335 gives dy k2 z − y z−y = = ρ where ρ = dz k3 a − z a−z Once again this equation is first-order linear. The 1 µ = e−ρ log(a−z) = (a−z) ρ . So the exact equation is

Taking the quotient of k2 . k3 integrating factor is

dy ρz dz ρy dz + ρ+1 = ρ+1 . (a − z)ρ (a − z) (a − z) Z Integrating, ρz dz y ρ = ρ+1 . (a − z) (a − z) To integrate the expression to the right of the equal, let u = a − z, a − u = z, and −du = dz. Integral becomes Z Z Z a du 1 a ρ a−u du = −ρ du + ρ = ρ− + C. −ρ uρ+1 uρ+1 uρ u ρ − 1 uρ−1 Back-substituting and multiplying through by (a − z)ρ , ρ a ρ ρ ρ y = a− (a − z) + C (a − z) = − + z + C (a − z) . ρ−1 ρ−1 ρ−1 Applying the same conditions as indicated on page 335 of a = 1, z = 0, and y = 0, we 1 . find C = − ρ−1

To find the equation for x in terms of z, we let σ = k1 /k3 and solve it the same way as we did for y, except that σ replaces ρ. y−x dx = σ . dz a−z 1 Replace y with the solution we got above for y. The integrating factor will be µ = (a−z) σ. You get Z − a + ρ z + C (a − z)ρ x ρ−1 ρ−1 dz + C 0 . σ = σ σ+1 (a − z) (a − z) Taking of the integral above is left to the reader.

S-43

Page 336 (§ 136. Simultaneous equations) example 1: This one can also be done using the Laplace method (see supplement pages S-31 through S-34). The equations are Dx + ay = 0,

and Dy + bx = 0.

In matrix form we have 

x

y

D b

a D



.

 √  √  The determinant of this matrix is D2 − ab = D − ab D + ab . Because there are no initial conditions given and forcing functions are zero, we can determine only the complementary function from this system of equations. That function is determined by the roots of√the determinant. If we apply these roots to x, the complementary function √ is x = C1 e abt + C2 e− abt . Substitute the derivative of this for dx/dt into the first equation to solve for y’s complementary function. Page 337 (§ 136. Simultaneous equations) example 2: Doing it using the book’s method, we take the derivative of the first equation, dx d2 x dy + =3 . 2 dt dt dt Subtracting the first equation from this gives d2 x dx dy dx − + −y =3 − 3x. dt2 dt dt dt Subtracting the second from the above gives dx d2 x dx − =3 − 4x. dt2 dt dt As indicated
in the text, this eliminates y and its derivative and we are left with 2 D2 − 4D + 4 x = (D − 2) x = 0. This gives a complementary function for x of C1 e2t + C2 te2t . Taking the derivative of this gives dx/dt = (2C1 + C2 ) e2t + 2C2 te2t . Substituting both of these into the first equation,
(2C1 + C2 ) e2t + 2C2 te2t + y = 3 C1 e2t + C2 te2t .

Solve for y to get the book’s complementary function for y.

Using the Laplace method we subtract 3x from both sides of the first equation and x from both sides of the second. This puts the equation in standard form with all dependent variables to the left of the equal. Forming the matrix from those equations yields x y   D−3 1 . −1 D−1

The determinant of this matrix is (D − 3) (D − 1) + 1 = D2 − 4D + 4. You can see how the same differential operator results from this method as from the book’s method. Apply this operator to x to get its complementary function, then use the substitution method shown above to find y’s complementary function from that of x. Page 337 (§ 136. Simultaneous equations) example 3: First using the book’s method. d2 x dy = µ , derivative of 1st equation 2 dt dt dy −µ2 x = µ 2nd equation times µ, but reversed dt d2 x + µ2 x = 0 difference eliminates y. dt2

S-44


Which is the same as D2 + µ2 x = 0. Observe the same operator evolves from using the matrix method. We rewrite the equations as dx/dt − µx = 0 and dy/dt + µx = 0. The matrix is x y   D −µ , µ D whose determinant is D2 + µ2 . If we ascribe to x the complementary solution, x = C1 cos µt + C2 sin µt, then dx/dt = −µC1 sin µt + µC2 cos µt. Substituting into the first equation, −µC1 sin µt + µC2 cos µt = µy, hence y = −C1 sin µt + C2 cos µt, and the relationships between the x-coefficients and the y-coefficients are as the book indicates. Page 337 (§ 136. Simultaneous equations) example 5: In the case where there are more than two dependent variables, the matrix method shows its superiority to the differentiate-and-substitute method. All three equations, as given, are in standard form – that is all of the dependent variables are to the left of the equal. The matrix, then, is x y z   D b c  a1 D c1  , a2 b 2 D

whose determinant is D3 − (a1 b + a2 c + b2 c1 ) D + a1 b2 c + a2 bc1 . This cubic is the same as the one shown in the book for this example, except using a different independent variable. Note that the book carelessly reused the symbol, z, as an independent variable for the cubic after it had already used it as one of the dependent variables in the original equation set. The two usages are not equivalent, and the book should have chosen a different symbol for the cubic to avoid confusion. Page 337 (§ 136. Simultaneous equations) example 6: In a modern electrical engineering class, this problem would almost certainly be solved using the Laplace method. We shall do it that way here. The matrix is i1  the determinant of which is

 

i2

MD

L2 D + R2

L1 D + R1

MD



 ,

M 2 D2 − (L1 D + R1 ) (L2 D + R2 )
= M 2 − L1 L2 D2 − (L1 R2 + L2 R1 ) D − R1 R2 .

Observe that the above quadratic is equivalent to the one given on page 338. The remainder of the problem – that is finding the particular solutions given that E1 and E2 are constants – is shown in the book. Page 339 (§ 136. Simultaneous equations, variable coefficients) example 3: To make the second solution work out, you have to multiply the equations by an integrating factor of x. dy dz x dx = = . 2 − y2 − z 2 x 2y 2z
Hence P = x2 − y 2 − z 2 /x, Q = 2y, and R = 2z. Multiplying these respectively by x, y, and z, and then taking the sum yields x2 + y 2 + z 2 . Hence by (3) on page 339, dy dz x dx + y dy + z dz x dx = . = = x2 − y 2 − z 2 2y 2z x2 + y 2 + z 2

S-45

The last expression is a perfect differential of 12 du/u, where u = x2 + y 2 + z 2 . Setting this equal to either of latter two expressions in the modified original equations (we’ll choose dz/2z) and integrating yields
1 2 2 2 1 = 2 log z + C, 2 log x + y + z

from which the book’s answer follows. Observe that you could have chosen dy/2y and
1 had 2 2 2 1 = 2 log y + C, 2 log x + y + z which would also be correct. Supplemental problem: If you simplify example 3 of page 339 by dropping z, you have the differential equation, dx dy − = 0. (a) x2 − y 2 2xy Based upon the solution for example 3 of page 339, we can infer that the solution to (a) is x2 + y 2 = Cy. Note that (a) is not exact. The problem then is, based upon the solution given here, to infer an integrating factor, µ (x, y), that makes (a) become exact. First we work the solution to (a) into a form in which taking the derivative eliminates the undetermined constant, C.
2x dx + 2y dy dy = log x2 + y 2 = log y + log C, whose derivative is , x2 + y 2 y Rearranging the derivative,   1 2x dx 2y dy = 0, + − x2 + y 2 x2 + y 2 y

which is exact because we arrived at it by taking the derivative of (a)’s solution. Summing the dy-fractions over a common denominator,

y 2 − x2 2x dx + dy = 0, x2 + y 2 y (x2 + y 2 ) which remains exact. In this form, it becomes clear how to form the integrating factor. To go from
(a) to the above, you would have to divide by x2 + y 2 , and multiply by
2x x2 − y 2 . Hence 2x x2 − y 2 µ (x, y) = . x2 + y 2

S-46


Page 340, apparent error in book: If u = f ax3 + by 3 , then applying the chain rule to take the partial derivatives of u with respect to x and y results in

∂u ∂u and = 3ax2 f 0 ax3 + by 3 = 3by 2 f 0 ax3 + by 3 , ∂x ∂y which differs from the partial derivatives shown in the book for this function. The book’s derivatives appear to be incorrect. Multiplying the partial (above) with respect to x by by 2 and the partial with respect to y by ax2 , we find that the right-hand side of both equations become equal. Hence ∂u ∂u − ax2 = 0, by 2 ∂x ∂y which also differs from the book’s conclusion on page 340. Page 341 (partial differential equations), example 1: Taking the partial derivatives with respect to x and y respectively, ∂u ∂u −b = b f 0 (bx − ay) and 1 − b = −af 0 (bx − ay) . ∂x ∂y Substituting for f 0 (bx − ay) from the first equation into the second, ∂u ∂u ∂u ∂u 1−b = a , hence a +b = 1. ∂y ∂x ∂x ∂y Page 341 (partial differential equations), example 2: Taking the partial derivatives with respect to x and y respectively,     1 ∂z 1 1 1 1 1 1 1 1 ∂z and − 2 . + 2 = 2 f0 − = − 2 f0 − − 2 z ∂x x x y x z ∂y y y x   Using the second equation to substitute into the first for f 0 y1 − x1 , 1 ∂z y 2 ∂z 1 = + , z 2 ∂x x2 x2 z 2 ∂y from which the book’s answer follows. −

hence

− x2

∂z ∂z + z 2 = y2 , ∂x ∂y

Page 341 (partial differential equations), example 3: This one is trivial compared with the preceding two examples. Both partial derivatives (with respect to x and with respect to y) are identically equal to a0 (x + y), and therefore equal to each other. So clearly the difference between the two partials must be zero, as asserted by the book’s answer. Page 344 (partial differential equations), example 3: The equation to solve is   ∂z a ∂z ∂z a ∂z a = z or equivalently + + = 1. ∂x ∂y z ∂x z ∂y

Make the substitutions, Z = log z, X = x/a, and Y = y/a. Then ∂Z a ∂z ∂Z ∂Z ∂Z a ∂z = and = ; equation becomes + = 1. z ∂x ∂X z ∂x ∂Y ∂X ∂Y In this case we have f (α, β) = α + β − 1 based upon the modified equation. Hence for f (α, β) = 0, we have β = 1 − α. The solution to the modified equation then is Z = αX + (1 − α) Y + C. Converting back to the original variables, 1 x y Z = log z = α + (1 − α) + C, hence z = Ce a {αx+(1−α)y} . a a

S-47

Page 345 (partial differential equations), example 4: The equation to be solved is    2  2  2 ∂z x ∂z y ∂z ∂z = z 2 or equivalently + y2 + = 1. x2 ∂x ∂y z ∂x z ∂y

If Z = log z, then dZ = dz/z, hence that substitution yields  2  2 ∂Z ∂Z x + y = 1. ∂x ∂y Similarly substituting X = log x and Y = log y, the equation becomes 2  2  ∂Z ∂Z + = 1. ∂X ∂Y

The √auxillary function for this modified equation is f (a, b) = a2 + b2 − 1 = 0. So b = 1 − a2 . By (1) on page 344, the solution is p p 1 − a2 log y + C. Z = aX + 1 − a2 Y + C hence log z = a log x + Taking the antilog of the latter equation results in the book’s answer.

Page 345 (partial differential equations,√type II), example 1: The book has definite errors on this problem. The equation, a2 + z · dz/dx = 2, is correct. From (2) on page 345, the solution is R√
3 a2 + z dz = 13 a2 + z 2 . x + ay + C = 12 Squaring both sides results in the solution,

(x + ay + C)2 =

1 9

a2 + z


3

,

(b)

which differs from the book’s answer (because the book misplaced a scalar). To show that this solution is correct, we substitute it back into the equation. Taking ∂/∂x of (b),
2 ∂z x + ay + C ∂z , hence 6 . = 2 (x + ay + C) = 31 a2 + z 2 2 ∂x ∂x (a + z)

By design, ∂z/∂y = a ∂z/∂x, which is easily confirmed by taking ∂/∂y of (b). Hence !2  2  2  2

∂z x + ay + C ∂z ∂z z + a2 . z+ = z + a2 = 36 2 2 ∂x ∂y ∂x (a + z) 2

Substituting from (b) for (x + ay + C) , 36

1 9

a2 + z

(a2


3

4

+ z)

z + a2




= 4,

which verifies that the solution is correct. Indeed if you replace value, the above test fails.

1 9

in (b) with any other

Page 345 (partial differential equations,
type II), example 2: Replacing q with bp, the equation becomes p 1 + b2 p2 = bp (z − a), which simplifies to √ 2 2 1 + b p = b (z − a). Solving for p we have p = φ (z) = 1b bz − ab − 1. By (2) on page 345, the solution is Z Z √ dz dz x + by + C = =b √ = 2 bz − ab − 1. φ (z) bz − ab − 1 The book’s answer follows from this. From that answer it’s easy to isolate z as a function of x and y. 1 ∂z 1 1 2 (x + by + C) + a + , hence = (x + by + C) . z= 4b b ∂x 2b

S-48

Substituting these results into 1 + b2 p2 = b (z − a), we have  2 ∂z 2 = 1 + 14 (x + by + C) 1 + b2 and b (z − a) = ∂x which verifies that the solution is correct.

1 4

2

(x + by + C) + 1,

Page 345 (partial differential equations, type III), example 2: Rearranging this we have p2 − x = y − q 2 = a. Now separating, it becomes p2 = x + a and q 2 = y − a. By (3) on page 345, the integral is Z Z √ √ z= x + a dx + y − a dy.

The book’s answer follows from taking the integrals. Observe that you could also have separated it into p2 = x − a and q 2 = y + a and obtained a slightly different solution. Explain how it is that these two solutions are, in fact, equivalent. Page 345 (partial differential equations, type III), example 3: Rearranging we have q/y = 2p2 = 2a2 . Notice that we set this to 2a2 rather than a in order to avoid a radical in the solution. You could set it to a and arrive at an equivalent solution. Separating it becomes q = 2a2 y and p = a. By (3) on page 345, Z Z z = a dx + 2a2 y dy = ax + a2 y 2 + C, which is the book’s answer.

Page 346 (equations analogous to Clairault’s), how to find singular solutions: The book offers no clue on how to do this. In this supplement beginning on page S-17, we developed the method that can be found in Ordinary Differential Equations and their Solutions by George M. Murphy (van Nostrand, 1960) for finding singular solutions to ordinary differential equations. That same method can be extended to apply to partial differential equations such as the ones found at the top of page 346. Murphy suggests that if p = dy/dx, then we take ∂/∂p of the entire equation, solve for p, then substitute that back into the original equation to obtain the singular solution. We now have the situation where p = ∂z/∂x and q = ∂z/∂y. The extended method is to take both ∂/∂p and ∂/∂q of the original equation, solve for p and q simultaneously, then substitute those solutions back into the original equations. Here is that method applied to the first example on page 346: ∂ (z = px + qy + pq) gives 0 = x + q, and ∂p ∂ (z = px + qy + pq) gives 0 = y + p. ∂q Clearly p = −y and q = −x. Substituting for p and q into the original equation, we have z = −xy − xy + xy = −xy, which is the singular solution given in the book. Applying the same method to the third example,   1 1 ∂ −1 z = px + qy − n (pq) n gives 0 = x − q (pq) n , ∂p   1 1 ∂ −1 n z = px + qy − n (pq) gives 0 = y − p (pq) n . ∂q

and

Solving for q in the first equation, q = xn pn−1 . Substituting q into the second equation, 1 n−1 1 
1 −1 −1 1−n = p (xn pn ) n = p (xp) , hence p = y 2−n x 2−n . y = p p xn pn−1 n 1

n−1

n

By symmetry we also have q = x 2−n y 2−n , and therefore pq = (xy) 2−n . The book’s singular solution follows by substituting these into the original equation.

S-49

The second example on page 346 is by far the most difficult on which to apply this method, but its singular solution also emerges in the same way.  p ∂  rp , (a) z = px + qy + r 1 + p2 + q 2 gives 0 = x + p ∂p 1 + p2 + q 2  p ∂  rq . (b) z = px + qy + r 1 + p2 + q 2 gives 0 = y + p ∂q 1 + p2 + q 2

From (a) we have x2 =

r 2 p2 , 1 + p2 + q 2

hence

x2 + p2 x2 + q 2 x2 = r2 p2

Solving for p2 ,
x2 1 + q 2 p = ; r2 − x2 2


and x2 + q 2 x2 = r2 − x2 p2 .

substituting p2 into (b), y 2 =

r2 q 2 1+

x2 (1+q2 ) r 2 −x2

+ q2

.

Now multiply top and bottom of the nasty fraction by r2 − x2 , expand the denominator, and then take the cancellations:

q 2 r2 − x2 r2 q 2 r2 − x2 2 , = y = 2 2 2 2 2 1 + q2 x + x + r − x2 q + q 2 r2 −  q 2 x from which


y 2 1 + q 2 = q 2 r2 − x2 , hence y 2 = q 2 r2 − x2 − y 2 . We easily find q from this, and by the symmetry of (a) and (b) we can find p as well: y x q = ±p and p = ± p . r2 − x2 − y 2 r2 − x2 − y 2

When these expressions are substituted for p and q in the original equation, we have (choosing the minus of the ±): s x2 + y 2 x2 + y 2 . z = −p +r 1+ 2 r − x2 − y 2 r2 − x2 − y 2 p Multiplying by r2 − x2 − y 2 and taking the cancellations, p z r2 − x2 − y 2 = −x2 − y 2 + r2 , from which the book’s singular solutions follows.

Page 348 (§ 141. Linear Partial Equations with Constant Coefficients) exam
ple 2: The book’s solution is in error. The operator equation is D2 − 4DD0 + 4D02 z = 0. 2 This factors into (D − 2D0 ) z = 0. If you substitute f (mx + y) into the equation, the 2 2 auxillary equation is m − 4m + 4 = (m − 2) = 0. Hence m = 2. By (9) on page 348, the solution is f (2x + y). Substituting this into the original equation verifies it as a solution, whereas the book’s answer fails to verify. Furthermore, the auxillary equation has a double root. We can’t just say that z = f1 (2x + y) + f2 (2x + y) is the general solution, since the f1 and f2 terms are equivalent. Equation (11) on page 349 shows how to deal with repeated roots. Accordingly the complete solution is z = f1 (2x + y) + xf2 (2x + y) . If you think about it, there is no reason that yf3 (2x + y) shouldn’t also be a solution. And indeed, if you substitute it into the original equation, you will see that it is. So why don’t we include it as a third term in the general solution? Because it is implicit in the existing two terms of that general solution. Consider that (2x + y) f (2x + y) is a single function of 2x + y, and is therefore a solution by being equivalent to the first term of the general solution. But the 2xf (2x + y) part of this is also a solution by being equivalent to the second term of the general solution. As such it becomes zero when

S-50

the equation’s operator is applied to it. That means that the yf (2x + y) part must also become zero when the operator is applied to it, so it must also be a solution. Hence it’s being a solution is implied by the general solution given above. So yf (2x + y), though a solution, is not independent of the general solution given above. Page 349 (§ 141. Linear Partial Equations with Constant Coefficients) example 1: Again the book gets the sign wrong. The auxillary equation factors into 2 (m + 1) . The double root is m = −1. The solution, z = f1 (x − y) + xf2 (x − y) follows from (11) on page 349. Observe that z = f1 (x − y) + yf2 (x − y) would also have been a suitable general solution. Page 349 (§ 141. Linear Partial Equations with Constant Coefficients) exam2 ple 2: The answer given in the book implies an auxillary equation of (m + 1) (m − 1) = 0. 3 2 This expands into m + m − m − 1 = 0. The partial differential equation that goes
with this auxillary equation is D3 + D2 − D − 1 z = 0. You will find that the book’s solution solves this equation and fails to solve the equation given in the example (although the f (x + y) term by itself does solve the book’s equation, as m = 1 is a root of m3 − 3m2 + m + 1 = 0). The book’s equation factors into   √  √  m − 1 m − 1 + 2 m − 1 − 2 = 0. Hence the general solution to the book’s equation (which has no repeated roots) is     √
√
z = f1 (x + y) + f2 1 − 2 x + y + f3 1 + 2 x + y .

Page 351 (§ 141. Non-homogeneous linear partial equations) example 3: Using the trial solution, z = eαx+βy results in the auxillary equation, αβ + aα + bβ + ab = 0,

which factors into

(α + b) (β + a) = 0.

The book’s solution follows from the paragraph following (12) on page 349. Page 351 (§ 141. Non-homogeneous linear partial equations) example 4: The equation factors into (D + D0 − 1) (D − D0 + 2) z = 0.

Again the book’s solution follows from the paragraph following (12) on page 349. Page 351 (§ 141. Non-homogeneous non-factorable linear partial equations) supplemental example: Some equations cannot be factored at all. This example is not in the book: ∂2z ∂2z − + z = 0. ∂x2 ∂y 2 αx+βy The trial solution, p z = e yields the auxillary equation, α2 − β 2 + 1 = 0, or equivalently, α = ± β 2 + 1. Substituting for α and inserting the undetermined scalars for each possible β, we have a solution of √ 2 X z= Cβ e± β +1 x+βy β

according to (15) on page 350. We can modify this solution by allowing β to assume purely imaginary values. So if β = ιω, then for ω ≥ 1,  p  X z= Cω sin ± ω 2 − 1 x + ωy + φω ω

is also a solution in accordance with Euler’s formula.

Example 1 on page 351 is also not factorable. Here the prototype solution was 2 z = eαx+α y . This one becomes of special interest when you allow α to assume the purely imaginary value of ιω.

S-51

Its general solution becomes z=

X

Cω e−ω

2

y

sin (ωx + φω ) .

ω

If you replace y with kt, where k is a heat conductivity rate, then this equation represents the temperature over time of a long uniform bar. Observe that if the temperature at t = 0 varies sinusoidally over the length of the bar, then the magnitude of that sinusoid decays in time at a rate in proportion to the square of its spacial frequency, ω. Diffusion of fluids in a long narrow tube complies with this same equation. Page 352 (Particular integrals of homogeneous partial equations) first example 2: The problem is to solve
−1 −1 −1 (x + y) = (D + 2D0 ) (D + D0 ) (x + y) . (a) D2 + 3DD0 − D02

Applying the right-hand factor first, we follow the procedure and let m = −1. Subtracting mx from y of the forcing function gives x + y + x = 2x + y. Integrating that dx gives x2 + xy. Now adding mx = −x to y of this result gives x2 + x (y − x) = xy. So at this point we’ve reduced the problem to finding −1

(D + 2D0 )

(xy) .

We repeat the same procedure, but this time with m = −2. Subtracting mx = −2x from y gives x (y + 2x) = xy + 2x2 . Integrating that gives 12 x2 y + 32 x3 . Now add mx = −2x to y of this result to get 21 x2 (y − 2x) + 23 x3 = 21 x2 y − x3 + 23 x3 , which is equal to the answer in the book. Alternatively we could have integrated (a) starting with the left-hand factor first. Then we have m = −2. Subtracting mx = −2x from y in (a) gives x + y + 2x = 3x + y. Integrating dx gives 23 x2 +xy. Now adding mx to y we have 32 x3 +x (y − 2x) = xy − 21 x2 . At this point we’ve reduced the problem to finding
−1 (D + D0 ) xy − 21 x2 .

We repeat the procedure, this time with m = −1. So subtract mx = −x from y to get x (y + x) − 12 x2 = xy + 21 x2 . Integrating this dx gives 12 x2 y + 16 x3 . Add mx = −x to y in this result gives 21 x2 (y − x) + 61 x3 = 21 x2 y − 21 x3 + 61 x3 , which is also equal to the book’s answer. Page 352 (Particular integrals of homogeneous partial equations) second example 2: Here F (D, D0 ) = D2 + 5DD0 + 6D02 , which has two factors, hence we will integrate twice. Dividing by D2 gives φ (D0 /D) = 1 + 5D0 /D + 6D02 /D2 . We also have b/a = 21 . So φ (b/a) = 1 + 25 + 64 = 5. So the particular integral will be ZZ 1 dx2 p (x, y) = , 5 y + 2x Z 1 = log (y + 2x) dx, 10 1 {(y + 2x) log (y + 2x) − y − 2x} . = 20

S-52

See S-67 for more on Fourier’s heat equation and its solution.

The 1902 edition of the book originally gave this example as
1 D2 + 5DD0 + 6D02 z = . (y − 2x)

In this case b/a is a root of φ (b/a). So the problem falls into the category explained at the top of page 353. Factoring this one, it becomes 1 , (D + 2D0 ) (D + 3D0 ) z = y − 2x   1 0 −1 0 −1 hence p (x, y) = (D + 2D ) (D + 3D ) . y − 2x

The factor that has the problem with b/a = − 12 is the left-hand factor. So we integrate just once using the right-hand factor. For that factor alone we have φ (b/a) = 1 + 3b a hence φ (b/a) = 1 − 32 = − 21 . So we take   Z dx −1 , p (x, y) = (D + 2DD0 ) −2 y − 2x −1

= (D + 2DD0 )

log (y − 2x) .

Now applying (18) on page 353 to the vanishing factor, we have p (x, y) = x log (y
− 2x). You are encouraged to substitute this expression for p into D2 + 5DD0 + 6D02 p and apply the operator to confirm that you get back 1/ (y − 2x). Page 353 (Particular integrals, case 3) first example 2: As the book suggests, we solve (D + DD0 − 2D0 ) z = sin (x − y) and (D + DD0 − 2D0 ) z = sin (x + y) separately, and then by the principle of superposition (see page S-24) we add the results to get the composite solution. Instead of using the method given on page 353, which, in this example, is prone to mistakes, we apply the method of undetermined coefficients (also introduced on S-24). In the case of the forcing function of sin (x − y), the particular integral must be a linear combination of sin (x − y) and cos (x − y). Hence (D + DD0 − 2D0 ) {λc cos (x − y) + λs sin (x − y)} = sin (x − y) .

Applying the operator to the expression in the curly-brackets,

λc {− sin (x − y) + cos (x − y)) − 2 sin (x − y)} + λs {cos (x − y) + sin (x − y) + 2 cos (x − y)} = sin (x − y) .

Separating terms and dividing out the sines and cosines yields −3λc + λs = 1,

λc + 3λs = 0,

3 hence λc = − 10 and λs =

1 10 .

Likewise we solve for the particular integral of sin (x + y) by solving (D + DD0 − 2D0 ) {λc cos (x + y) + λs sin (x + y)} = sin (x + y) .

Applying the operator to the expression in the curly-brackets,

λc {− sin (x + y) − cos (x + y)) + 2 sin (x + y)} + λs {cos (x + y) − sin (x + y) − 2 cos (x + y)} = sin (x − y) .

Separating terms and dividing out the sines and cosines yields λc − λs = 1,

Hence λc =

1 2

and λs =

p (x, y) =

3 − 10

− 12 .

−λc − λs = 0. So the complete particular integral is

cos (x − y) +

1 10

sin (x − y) +

1 2

cos (x + y) −

1 2

sin (x + y) ,

which differs from the book’s answer. And by the way, did you try to find the complemen-

S-53

tary solution to this one? If z = eαx+βy , then the auxillary equation is α + αβ − 2β = 0. A few more algebraic steps gets you to β o 1+β n X z= . Cβ e2x+(1+β)y β

Now that we have solved the more difficult problem, I’ll reveal that I believe that the typographical error in the book was not in the
answer given, but in the original equation, which should have been D2 + DD0 − 2D02 z = sin (x − y)+ sin (x + y). The particular integral for this one is easily solved using the method in case 3 of page 353. 1 1 sin (x ± y) = sin (x ± y) . D2 + DD0 − 2D02 −1 ∓ 1 + 2 Observe that by using the ± notation, we solve both cases at the same time. In the − case of the ±, we immediately see that we get 12 sin (x − y). In the + case, the denominator vanishes. But the operator factors into D + DD0 − 2D2 = (D + 2D0 ) (D − D0 ) .

So by case 2 and (18) on page 353, with a = 1 and b = 1, we have Z 1 0 −1 0 −1 (D + 2D ) (D − D ) sin (x + y) = x sin (x + y) dx = − 31 x cos (x + y) . 1+2 Combining the two results still differs from the book’s answer by the sign of the cosine term. You are encouraged to apply the original operator to the expression above to see which is the right answer. The complementary solution to this equation is f1 (2x − y) + f2 (x + y) in accordance with (9) on page 348. So the complete solution is z = f1 (2x − y) + f2 (x + y) +

1 2

sin (x − y) − 13 x cos (x + y) .

Page 353 (Particular integrals, case 4) second example 1: We have a = 2 and b = 3. Accordingly if F (D, D0 ) = D2 − DD0 − 2D02 + 2D + 2D, then F (a, b) = a2 − ab − 2b2 + 2a + 2b = 4 − 6 − 18 + 4 + 6 = −10. The book’s answer follows. Page 353 (Particular integrals, case 4) second example 2: We have F (n, m) = nm + an + bm + ab = (n + b) (m + a). The book’s answer follows. Page 354 (Particular integrals, case 5) second example 2: First divide out 3D from the operator:  −1   −1 1 D D02 D D02 D0 1 D0 1− − −1+ xy = − − + xy. 3D 3 3D D 3D 3 3D D No sooner does the book state, “The expansion is not usually carried higher than the highest power . . . in f (x, y),” than it gives an example that requires carrying out the expansion higher than that. In this case we need to take the expansion out to the cubed term. The reason is the D that appears in the denominator of terms that will be in the expansion. You have to use your careful judgement on this. Look at how the powers expand. Any D or D0 to the squared or higher power will cause xy to become zero. Hence the D02 /3D term in the above contributes nothing to the solution and can be ignored throughout your analysis. There is another very sneaky aspect to this problem that I will show as we analyze the expansion. The expansion out to the cubed term is then ( )    2  3 D D02 D D02 D0 D0 D0 1 D D02 1+ + − + − + − + − + + . . . xy. 3D 3 3D D 3 3D D 3 3D D Going through term by term, we see that 1 xy contributes − 3D

S-54

− 61 x2 y.

Next, recalling that the D02 /D contributes nothing, we have,   D D02 D0 1 − + xy which contributes − 19 xy − − 3D 3 3D D

Next, again ignoring the D02 /D term,  2 D D02 D0 − + expands to 3 3D D

1 3 18 x .

2D0 D02 D2 + + 2, 9 3 D

Only the middle term has powers of D or D0 less than squared, so it is the only term that contributes. Hence 1 2D0 xy contributes − 91 x2 . − 3D 3 Finally, again ignoring D02 /D, 3  D0 DD0 D02 D03 D D02 D3 − + + + + 3, expands to 3 3D D 27 3 D D

of which only the DD0 /3 term will make any contribution. This expansion is in accordance with the binomial theorem, but here is where there is a trap for you. The binomial theorem assumes multiplication to be commutative. But in this case that assumption leads to a wrong answer. Letting a = D/3 and b = D0 /D, we have, (a + b) (a + b) (a + b) = a3 + aab + abb + aba + baa + bab + bba + b3 .

But notice that if you apply aa to xy before applying b to it, you get zero. The order of application here does matter. Hence only two out of the three terms containing two a’s and one b contribute anything. The one that applies both a’s before applying the b contributes nothing. So the real contribution of the cubed expansion is 1 2DD0 2 − xy which contributes − 27 x. 3D 9 You should confirm for yourself that expanding powers higher than cubed leads to all of the terms containing a factor of Dn or D0n where n ≥ 2. So no higher powers are needed. Gathering all of the contributions shown above and summing them yields the book’s answer. You are encouraged to substitute that answer back into the original equation to confirm that it does work. Note that you could have started by dividing by 3D0 rather than 3D. Using the 1 3 2 same procedure as above, you would have arrived at 16 xy 2 + 18 y + 19 xy + 19 y 2 + 27 y. This solution also works. If p is the book’s answer and q is this answer, then Cp + (1 − C) q also solves the particular integral. Page 355 (nonconstant coefficients) first example 2: Making the substitution, the equation becomes ∂2z ∂z ∂z ∂2z ∂2z − +2 + 2− = 0. 2 ∂u ∂u ∂u∂v ∂v ∂v Letting D and D0 be partial derivatives with respect to u and v respectively, we have
D2 − D + 2DD0 + D02 − D0 z = 0.

The auxillary equation is

α2 − α + 2αβ + β 2 − β = (α + β)2 − (α + β) = 0.

Solving we find that either α + β = 0 or α + β = −1. The first root leads to the solution, z = g1 (v − u), in accordance with (9) on page 348. The second root leads to the solution, z = eu g2 (v − u) in accordance with B on page 349. So the complete solution is z = g1 (v − u) + eu g2 (v − u) . u Back-substituting x = e and y = ev , and allowing fk (s) = gk (es ) for both functions yields the book’s answer.

S-55

Page 355 (nonconstant coefficients) example 3: Making the suggested substitution, we have ∂z ∂2z dv v= and = . ∂x ∂x∂y dy Equation becomes dv − av = 0. (x + y) dy Separating variables dy dv a = . x+y v Integrating gives a log (x + y) + g (x) = log v

or equivalently

a

f1 (x) (x + y)

= v,

g

where f1 = e . Back-substitute for v and we have a

f1 (x) (x + y)

=

∂z . ∂x

Integrating dx yields the book’s answer. Page 355 (nonconstant coefficients) second example 2: Expanding the book’s formulation in ϑ, or equivalently

ϑ2 − (1 + n) ϑ + 2ϑϑ0 + ϑ02 − (1 + n) ϑ0 + n = 0, 2

(ϑ + ϑ0 ) − (1 + n) (ϑ + ϑ0 ) + n = 0. This is a quadratic in ϑ + ϑ0 , and is easily factored into (ϑ + ϑ0 − 1) (ϑ + ϑ0 − n) = 0.

Allowing u = log x and v = log y, then by B on page 349,

z = eu g1 (v − u) + enu g2 (v − u) ,

from which the book’s answer follows by back-substituting x = eu , y = ev , and fk (s) = gk (es ). Page 356 (Solutions in power series) Jumbled equation: In the uniqueness proof, it should go like this: If y = φ (x) is the power series solution to dy/dx = y, and y = vφ (x) were a different solution, then dy − y = v 0 φ (x) + vφ0 (x) − vφ (x) = 0. dx But since φ (x) solves the differential equation, that means φ0 (x) = φ (x). Hence the above equation collapses into dy − y = v 0 φ (x) = 0, dx requiring either v 0 or φ (x) to be identically zero. So if φ is not zero, v must be a constant. This same strategy can demonstrate uniqueness of the general solution to any first order linear equation that has no forcing function. If y = φ (x) solves the equation, dy + p (x) y = 0. (a) dx We suppose that y = vφ (x) also solves (a). Then dφ + v 0 φ (x) + p (x) vφ (x) = 0. (b) v dx But since φ (x) solves the original equation, we have dφ = −p (x) φ (x) . dx Substituting the above into (b) yields v 0 φ (x) = 0, which is the same result as before.

S-56

Page 357 (solutions in power series) example 4: The method given in the book for determining series solutions is also known as the method of Frobenius, after mathematician, Ferdinand Georg Frobenius (1849-1917). The modern book, Mathematical Methods in the Physical Sciences, 2nd edition, by Mary L. Boas, (1983, Wiley and Sons) provides an especially lucid presentation of this method in chapter 12. The problem at hand is to develop a series for the solution to d2 y + xy = 0. dx2 The table below breaks the series solution into its elements according to the terms of the above equation. xm xm+1 xm+n d2 y dx2 xy

(m + 2) (m + 1) a2

(m + 3) (m + 2) a3

(n + m + 2) (n + m + 1) an+2

a0

an−1

Each row of the above table is the contribution of one of the terms of the differential equation. Each column shows the contributions attributed to each power of x by various terms of the equation. The quantity, m, is the exponent of the lowest nonzero term of the series solution (note that in general m is not necessarily positive and not necessarily an integer). We use the right-most column of the table to determine the recursion relationship among the an ’s. The sum of that column must be zero in order to satisfy the original equation. Hence −an−1 an−1 +(n+m +2) (n+m+1) an+2 = 0 consequently an+2 = . (n + m + 2) (n + m + 1) Replacing n with n + 1, we have −an an = − (n + m + 3) (n + m + 2) an+3 consequently an+3 = . (n + m + 3) (n + m + 2) To determine legitimate values for m, we use the left-hand equation above to run the the recursion backward. At the point where n < 0, we need to get a zero result in order to prevent the series from continuing indefinitely into negative indices. Note that the recursion relationship requires that n − m be a multiple of 3. This means that when n − m = −3, we have 0 = m (m − 1). Hence m = 0 or m = 1. Running the recursion forward with m = 0 and a0 = 1, we see that 1 1 1 a0 = 1; a3 = − ; a6 = ; a9 = − ; ... 2·3 (2 · 3) (5 · 6) (2 · 3) (5 · 6) (8 · 9) Doing the same with m = 1 and a1 = 1, we see that 1 1 1 ; a7 = ; a10 = − ; ... a1 = 1; a4 = − 3·4 (3 · 4) (6 · 7) (3 · 4) (6 · 7) (9 · 10) You should be able to see that these coefficients are equal to the ones given by the book’s answer. The two series (where m = 0 and where m = 1) are independent of each other, so the general solution is a linear combination of those two series. It is useful to see what happens when we allow m = 2. That means that a2 , a5 , a8 , . . . would all be nonzero. Hence there is a nonzero term in the series, a2 x2 . The second derivative of this term is the constant, 2a2 . That means that some other term of the series times x must cancel 2a2 . But that term would have to come from a−1 x−1 , which, by assumption, is zero. So if we were to allow nonzero a2 , then the series would continue with nonzero terms indefinitely into negative indices. You are encouraged toPexperiment for yourself with applying the differential equation to a series consisting of a3n+2 x3n+2 to see for yourself why it can never work unless all of these terms are zero.

S-57

It is fairly easy to prove that both the m = 0 and the m = 1 series converge for all x. Each nonzero term of either series is arrived at by multiplying the last nonzero term by x3 and dividing it by (n − 1) n or by n (n + 1), where n is the index of the new term. Clearly you can always find large enough n that the divisor will exceed x3 , no matter large x is. So for large enough n, the series will always pass the ratio test and therefore must converge. This convergence analysis applies to this example only. There are other differential equations for which the series solutions have finite radii of convergence. Returning now to the problem of solving for possible values of m, there is an abbreviated method for doing this once you have established the table. Usually all non-blank rows the left-most column of the table will have only a single index for a. If so, take the sum of all rows in that column. In the case of this example, only the first row of the left-most column is non-blank. It shows an expression involving a2 , namely (m + 2) (m + 1) a2 . Let k be the index of a, in this case, k = 2. The procedure is to replace m with m − k, drop the a, and set the resulting expression to zero. In this case you would have (m + 2 − k) (m + 1 − k) = m (m − 1) = 0.

(c)

Equation (c) is known as the indicial equation. Simply solve it for m. In this case it is clear that m = 0 or m = 1. For this example the solution is easy and results in m having two nonnegative integer values. More complicated examples may have more than one row of the left-most column of the table containing nonzero entries. Solving for m will then involve finding the roots of an indicial polynomial whose degree is equal to the order of the original equation. Those roots will not necessarily be integer or positive. Indeed they could even be non-real complex complements. Whatever they be, each such solution for m will be the basis for an independent series solution for the differential equation in question, and will appear as a summand in each exponent. As such, it will always be possible to factor xm from the series solution to arrive at a solution in the ∞ ∞ form of X X y= an xm+n = xm an xn . n=0

n=0

Page 358 (Expansion of J0 into a power series): Substituting n = 0 into the Bessel equation given on page 358 gives

1 dy dy d2 y d2 y + + y = 0 or equivalently x 2 + + xy = 0. (d) 2 dx x dx dx dx The equation on the left is a canonical form, d2 y/dx2 + P (x) dy/dx + Q (x) y = 0, that is useful in the analysis of second order linear equations. We see that in the case in which we are interested, P (x) = 1/x and Q (x) = 1. We shall come back to that later, but for the purposes of finding a series solution, J0 (x), to this, we shall start with the right-hand form of (d). We could just go ahead and assemble the table from that form, which would yield the correct series, but doing it that way would hide an important piece of information. Instead we observe that   dy dy d d2 y x . = x 2+ dx dx dx dx Making that substitution, (d) becomes   dy d x + xy = 0. dx dx

This is the form out of which we build the table. If yn (x) = an xn , then x yn (x) = an xn+1 , and     d dyn d dyn+1 2 2 n−1 x = an n x hence x = an+1 (n + 1) xn . dx dx dx dx

S-58

(e)

Based upon those facts, we fill out the table as follows: xm

xm+1

xm+n

xy a0 an−1   dy d 2 2 2 x a1 (m + 1) a2 (m + 2) an+1 (m + n + 1) dx dx Using the method we used on the previous example, the indicial polynomial is m2 = 0. Observe that this has a double root at m = 0. So we cannot come up with two independent series arising from the two solutions for m. Instead we will develop the series for just one of them, using m = 0, and make some comments on the other. The right-most column of the table shows the recursion to be an an−1 + an+1 (m + n + 1)2 = 0 consequently an+2 = − (m + n + 2)2 Because m = 0 and the recursion goes in steps of two, we get a series with only even powers of x. Allowing a0 = 1, we have ∞ 2n X x4 x6 x2 n x . + − + · · · = (−1) J0 (x) = 1 − 2 2 2 22n n!2 (2) (2 · 4) (2 · 4 · 6) n=0

Given that Γ (1) = 1, this is the same as the series given in on page 358 with n = 0. Using the same type of argument as in the previous example, we see that this series converges for all x.

We return to the canonical form of the original equation. Recall that the canonical form is d2 y/dx2 + P (x) dy/dx + Q (x) y = 0. In the case of J0 , we have P (x) = 1/x and Q (x) = 1. The series development we just did results in a Taylor series taken around x = 0, which is precisely the point at which P (x) is discontinuous. Compare that with example 4 on page 357, which is already in canonical form with P (x) = 0 and Q (x) = x. In that example, both coefficient functions are continuous and differentiable everywhere. The distinction has a bearing on how we use the roots of the indicial equation. This topic is quite involved, and we can only touch on it here. Ordinary Differential Equations and Their Solutions by George M. Murphy (van Nostrand, 1960) goes into full detail in part I, unit B1. Other advanced books on differential equation also deal with these issues. When P and Q are continuous and differentiable at the point around which you are taking the series, then you can use both roots of the indicial equation as we did in example 4 on page 357, provided the roots are distinct. If, however, there is a pole in either P or Q at the center-point of the series, as we have in the equation for J0 , the situation becomes more complicated. In that case, if the two roots differ by an integer, then the two series developed from those two roots will not be independent of each other. This happens when you develop the series for Jn , where n is any integer other than zero. The roots of the indicial equation are m = ±n. The series from the root, m = −n, has negative powers of x. The denominators of those terms contain a factor of Γ (k), where k is a nonnegative integer. But the Γ function of nonnegative integers is infinite, so those terms of the series are all zero. The terms that remain are all a scalar multiple of those of the other series developed from the root, m = n. Hence the two series are dependent. In addition when you have a double root in the indicial equation, as we do for J0 , we also have no way of developing two independent series. We can use the formula on page 307 to establish a second independent solution when we already know a particular solution. In the of the zeroth order Bessel function, we know that J0 (x) solves equation (d). Using P (x) from the canonical form, we have Z R 1 −P (x)dx dx, Y0 (x) = J0 (x) 2 e J0 (x)

S-59

where Y0 (x) is the second independent solution to (d). Since P (x) = 1/x, the integral Z becomes dx Y0 (x) = J0 (x) 2. x J0 (x) Observe that J0 (x) is nonzero at x = 0. Based upon the series solution for J0 (x), it is clear that 1/J0 (x)2 and all its derivatives exist at x = 0. This means that 1/J0 (x)2 can be expanded as a Taylor series around x = 0, and that series will have a radius of convergence equal to the lowest magnitude x that solves J0 (x) = 0. Carrying out that expansion is beyond the scope of this tutorial. It is clear, however, that the constant term of that expansion will be equal to 1. That term divided by the x that appears inside the integral is responsible for J0 (x) log x term that appears in the expansion for Y0 (x). The first few terms of this expansion is shown on page 358, although the book uses the old symbol, K0 , for this function rather than the modern symbol, Y0 . Developing a series solution around points other than x = 0: That is, if we seek a series around x = b, for example, we are looking to establish the ak ’s in ∞ X 2 3 n y (x) = a0 + a1 (x − b) + a2 (x − b) + a3 (x − b) + . . . = an (x − b) . n=0

The method here is a straightforward variation on the method of Frobenius already introduced on page S-57. Simply make the substitution, z = x − b, into the original differential equation. As an example, the zeroth order Bessel equation, (d), becomes   dy 1 dy d d2 y d2 y z + (z + b) y = 0. + + + y = 0 or equivalently b dz 2 z + b dz dz 2 dz dz Develop the series solution for z around z = 0 using the methods already presented, then back-substitute at the end. In the case of the zeroth order Bessel function, the series is more complicated than any we have attempted. We will only go as far as to fill out the table for this example. zy by   d dy z dz dz

zm

z m+1

z m+n

a0 b

a0 a1 b

an−1 an b

2

a1 (m + 1)

2

a2 (m + 2)

2

an+1 (m + n + 1)

d2 y a2 b (m+2) (m+1) a3 b (m+3) (m+2) an+2 b (m+n+2) (m+n+1) dz 2 To form the indicial equation, sum up the left-most column. Observe that we have mixed indices of a. Let k be the highest of those indices – in this case k = 2. Replace m with m − k. The resulting equation is b

2

a−2 b + a−1 (m − 1) + a0 b m (m − 1) = 0.

By assumption, a0 is the lowest index coefficient that is nonzero. It follows that a−2 and a−1 are both zero. So the indicial equation is m (m − 1) = 0. Summing the right-most column, though, we see that the recursion is nasty and involves four different indices of a. In this example there is no clear path to a closed form for an . Were we able to establish an , however, we could not expect that either the m = 0 or the m = 1 series to be equal to either J0 or Y0 . Instead each would be some linear combination of the two. And because Y0 has a pole at z = −b, we would expect that the radius of convergence of either series would be b.

S-60

Resonant systems revisited: On page S-26 is a paragraph introducing the canonical form of the equation for resonant systems. Resonant systems occur again and again in engineering. Here are three physical examples. i. An RLC-circuit. The diagram shows such a circuit. A voltage, vin , which varies sinusoidally in time, is applied on the left and results in an output voltage, vout , on the right. Here vin is the forcing function and vout is the particular integral deriving from the differential equation implicit in the circuit acting upon the forcing function. We let vin = cos ωf t, where we shall call ωf , the forcing frequency. When ωf = 0, the impedance of the capacitor, C, is infinite and the impedance of the inductor, L, is zero. So we expect then that vout = vin . As ωf goes to infinity, the impedance of the inductor goes to infinity as well while the impedence of the capacitor goes to zero. Under those conditions we should expect that vout should approach zero. Analysis of this circuit network yields 1 LC vin . vout = 1 R D2 + L D + LC The analysis is based upon the impedance of the inductor being LD, the impedence of the capacitor being 1/ (CD), and the impedence of the resistor being R. r r 1 C 1 and ζ = R , we find that Letting ω = LC 2 L ω2 vin , vout = D2 + 2ζω D + ω 2 which is similar to the resonant equation on page S-26, where ω is the resonant frequency and ζ is the damping factor. Substituting vin = cos ωf t, this becomes ω2 cos ωf t. (a) D2 + 2ζω D + ω 2 The goal is to see how vout varies with the forcing frequency, ωf . We can solve for the particular integral, vout , using the method detailed in case 5 on page 313. Replacing D2 with −ωf2 , ω2  cos ωf t.  vout = 2ζω D − ωf2 − ω 2   Now multiply top and bottom by 2ζω + ωf2 − ω 2 to get o n  2ζωD + ωf2 − ω 2 ω 2 vout = 2 cos ωf t.  4ζ 2 ω 2 D2 − ωf2 − ω 2 vout =

Again replacing D2 with −ωf2 , vout

o n  2ζωD + ωf2 − ω 2 ω 2 = − 2 cos ωf t.  4ζ 2 ω 2 ωf2 + ωf2 − ω 2

To simplify, divide top and bottom by ω 4 and replace ωf /ω with the frequency ratio, σ.

2 ωζ D + σ 2 − 1 1 − σ 2 cos ωf t + 2ζσ sin ωf t vout = − . (b) 2 cos ωf t = 2 4ζ 2 σ 2 + (σ 2 − 1) (1 − σ 2 ) + 4ζ 2 σ 2 It is possible to resolve this function into two components. One is the amplitude of the

S-61

resulting sinusoid. To find that we take the square root of the sum of the squares of the coefficients of cos ωf and sin ωf .

A plot of this function is shown to the right for various values of ζ. This type of plot showing the magnitude of output of a system as a function of sinusoidal frequency of input is known as a Bode plot, after Hendrik Wade Bode (19051982), who invented it. Note that the axes of the plot are scaled by the log base 10 of frequency and the log base 10 of amplitude.

kvout k 1 . =q kvin k (σ 2 − 1)2 + 4ζ 2 σ 2

(c)

Based upon this plot we can see that, as expected, for ωf  ω (that is for 0 < σ  1), we have kvout k/kvin k ≈ 1, which is consistent with our early observation that vout = vin when ωf = 0. When ωf  ω, (that is when σ  1), we see that on the log/log scale, the plot decreases with a slope of −2. This is equivalent to kvout k varying inversely as σ 2 , which is consistent with our early observation that as ωf goes to infinity, we expect vout to go to zero. Of course the region of greatest interest is near the resonant frequency – that is the region around σ = 1, or equivalently the region around ωf = ω. When ζ < 1, we see that the plot reaches a maximum just to the left of σ = 1. As ζ gets closer and closer to zero, the maximum becomes sharper and sharper, as well as becoming higher. The position of the maximum approaches σ = 1. Observe that at ζ = 0.1, we have the amplitude of vout being five times the input voltage at the point of resonance. Indeed, when σ = 1, equation (c) predicts that kvin k . 2ζ So as ζ approaches zero, at resonance the ratio of the output amplitude to the input amplitude goes to infinity. Circuits such as this one designed to have a sharp resonance (that is a very small value for ζ) are useful in radio receivers, among other applications, where it is necessary to select a narrow band of frequencies out of a broader spectrum input. kvout k =

By contrast, look at what happens when ζ is much greater than 1. The trace for ζ = 5 shows that for ωf up to about 0.1ω, we continue to have kvout k ≈ kvin k. From about 0.1ω to about 10ω, the trace decreases with a slope of −1. Only for ωf > 10ω do we see the decrease rate become a slope of −2. What is happening here is that when ζ > 1, the denominator of equation (a) will have two real roots, one√of them with |D| < ω, the √ other

with |D| > ω. In the case of ζ = 5, we see D = −5 + 24 ω and D = −5 + 24 ω, or D ≈ −0.1ω and D ≈ −10ω. When −ωf falls between those two roots, the factor

S-62

(D + 0.1ω) is growing proportionally larger, whereas the factor (D + 10ω) is changing very little. So between the two roots, the amplitude varies in inverse proportion to the forcing frequency. But when the forcing frequency is larger than the magnitude of both the roots, the amplitude varies in inverse proportion to the square of the forcing frequency. Also observe that as ζ becomes larger and larger, the two roots will spread farther and farther apart (although both roots will always be negative). Recalling that the method of finding the particular integral given on page √ 313 replaces D2 with −ωf2 , this means that it also replaces D with ι ωf , where ι = −1. If the two roots are at −0.1ω and −10ω, then the contribution to the denominator of (a) of the two roots will be ι ωf + 0.1ω and ι ωf + 10ω respectively. The use of complex quantities in the analysis of linear circuits is commonplace in the world of electrical engineering. Indeed if all you had done to equation (a) was replace D with ι ωf and evaluated the resulting complex function, the real part of the result would give you the correct cosine coefficient and the imaginary part the correct sine coefficient. From those coefficients the right-hand version of equation (b) would follow√ immediately. And just so you know, the customary symbol in electrical engineering for −1 is j,which distinguishes it from i, which electrical engineers use as a variable name for electrical current. Electrical engineers also use the symbol, s, instead of D. The discussion of equation (b) already stated that it has a second component besides its magnitude. That second component is its phase angle. When you add the sine and cosine terms of equation (b), you get a sinusoid that lags in phase from vin . The plot here shows that phase difference as a function of σ = ωf /ω for various values of ζ. We can see that when the forcing frequency is substantially less than ω, there is very little phase lag between the output and the input. At the resonant frequency, the output always lags the input by exactly 90◦ . And as the forcing frequency becomes much greater than the resonant frequency, the phase lag approaches 180◦, which means that the vout is almost completely opposite to vin . The other trend to notice is that as ζ approaches zero, the phase lag transition from nearly 0◦ to nearly 180◦ occurs more and more sharply right at the resonant frequency. The function used to construct the above plot is: 2ζσ , 1 − σ2 then summing in 180◦ where the arctangent was negative to account for quadrant. ∠vout |vin = tan−1

If you had processed equation (a) by replacing D with ι ωf , then you would have had a point on the complex plane for each value of ωf . Drawing a ray from the origin through that point would make an angle with the real axis that is exactly equal to the phase lag of vout to vin corresponding to that ωf . In addition, the distance of that point from the origin would be exactly equal to the corresponding magnitude, kvout k/kvin k.

S-63

As promised earlier, there are other examples of physical systems that follow the resonance equation, (a). ii. An automobile’s suspension: The axle of an automobile is connected to the chassis by a spring and a dashpot (which mechanics call a shock-absorber) in parallel. Let the chassis have mass, m, and let xin be the vertical position of the axle and xout be the position of the chassis. The spring applies a force, k (xin − xout ) to the chassis.
The dashpot applies a force, c dxdtin − dxdtout to the chassis. Be aware that this is a very simplified description of an automobile suspension. Other factors in real suspensions are couplings between the four wheels, that the suspension is responsive not only in the vertical axis but the lateral and longitudinal axes as well, and a number of other engineering details. But the single-wheel, single-axis system described here will show how resonant systems occur in mechanics. By Newton’s laws of motion, we have ftotal = k (xin − xout ) + c



dxin dxout − dt dt



= m

d2 xout , dt2

(d)

or equivalently

k (xin − xout ) + c (xin − xout ) D = m xout D 2 . Solving for xout in terms of xin ,

xout

If we let ω =

r

k c + D m m xin . = k c 2 D + D+ m m

1 c k and ζ = √ , the equation becomes m 2 km xout =

ω 2 + 2ζωD xin . D2 + 2ζωD + ω 2

Finally we let xin = cos ωf t: xout =

ω 2 + 2ζωD cos ωf t, D2 + 2ζωD + ω 2

(e)

which is similar to equation (a), in that it has the same denominator, but is different in that it has an extra term in the numerator. We will solve this for xout by substituting D with ιωf , as suggested on the previous page. ω 2 + 2ζιωf ω cos ωf t, 2ζιωf ω − (ωf − ω 2 )
Multiply top and bottom by 2ζιωf ω + ωf − ω 2 , then divide top and bottom by ω 4 and replace ωf /ω with the frequency ratio, σ, take the cancellation, rearrange a bit, and you have
4ζ 2 σ 2 − σ 2 − 1 − 2ζισ 3 cos ωf t. (f) xout = 4ζ 2 σ 2 + (σ 2 − 1)2 xout =

S-64

A plot of the magnitude of kxout k/kxin k for various values of ζ is shown to the right. Observe that like Bode plot of equation (c), for ωf  ω, the output magnitude is nearly equal to the input magnitude. The region around the resonance point is nearly identical to that of (c). But for ωf > ω, the behavior is different. For low values of ζ the trace slopes down steeply after resonance. For high values of ζ, the trace doesn’t even start to slope down appreciably until well to the right of resonance. For all values of ζ, the slope far to the right of resonance eventually approaches −1. This of course begs the question of what value of ζ should be chosen for an automobile suspension. The goal is that the suspension will isolate the chassis from irregularities in the road. So for high values of ωf , which correspond to rapid changes in the road surface, we would like to see them well below the 1.0 line in the plot. For low values of ωf we would like the chassis to follow the axle. A low value for ζ does that the best except for the region right around resonance. In that region a low ζ amplifies those frequencies, and the effect will be that the car bounces at the resonant rate each time it goes over a bump. So the actual value of ζ chosen by automotive designers is a compromise. Typically the suspension is designed for ω ≈ 2π radians/second and with ζ between 0.5 and 1.0. iii. A servo-system: A potentiometer is arranged in a circuit so that its wiper-pin produces a voltage that is in proportion to how far the potentiometer shaft is turned. That voltage goes to an amplifier that delivers a current to a servomotor in proportion to the input voltage. The servomotor delivers a torque to a flywheel in proportion to its input current. The flywheel also has a mechanical damper that provides a resistive torque in proportion to how fast the flywheel spins. Finally, the shaft of the motor is coupled to the shaft of a second potentiometer arranged in the same way as the first. The wiper-pin voltage of that potentiometer is subtracted from the wiper-pin voltage of the first, and it is actually the difference of the two voltages that is the input of the amplifier. The desired effect is that if an operator turns the shaft of the first potentiometer by some angle, the servo-system will turn the flywheel by the same angle, after which the flywheel will come to rest. Or more broadly, we desire the flywheel’s angular position to follow that of the first potentiometer shaft. Let k be conversion factor that relates servomotor torque to potentiometer position difference. Let c be the resistive factor that relates resistive torque on the flywheel to the angular velocity of the flywheel. Let m be the moment of inertia of the flywheel. Then the equation descibing the whole servosystem is identical to (d) in the previous

S-65

example, where xin is the angular position of the first potentiometer shaft, xout is the angular position of the flywheel, and ftotal is replaced by total torque. Observe that for low values of ζ, the flywheel will overshoot, then oscillate around the control point before settling, each time the first potentiometer is moved. For high values of ζ, the flywheel response will be sluggish, undershooting the first potentiometer’s change of position and settling into the control point only over time. Again the ideal choice for ζ is a compromise value between 0.5 and 1.0. The the galvanometer on page 323 is yet another example forcing an input on a resonant system. Differential Equation Solution to Keplerian Orbits: Here we solve the problem of two bodies attracted to each other by gravity. In particular we solve the problem where one of the two bodies outweighs the other by a large factor (for example, the sun’s mass is over 300,000 times the mass of the earth). This way we can consider the larger mass as fixed and solve only for the orbit of the smaller mass, m. We set the problem up in polar coordinates. In the radial axis we have  2 d2 r dθ m 2 − mr = −f (r) . dt dt

(a)

Here f (r) is the attractive gravitational force felt by the smaller object as a function of how far it is from the larger object. The first term on the left is simply Newton’s law of motion. The second term is the so-called centrifugal force. That is it is the apparent force a rotating object feels away from the center of rotation. The gravitational force acts only in the radial direction. Hence in the tangential axis we have conservation of angular momentum. Angular momentum, L, is given by dθ (b) L = mr2 . dt By saying that this quantity is conserved, we are saying that it is constant and therefore its time derivative is zero.   d d2 θ dr dθ 2 dθ m r = 0, or equivalently r 2 + 2 = 0. dt dt dt dt dt Observe that geometrically, angular momentum, L, is the mass, m, times twice the rate of radial area swept by the smaller object as it orbits the larger object. Johanas Kepler observed that planets orbiting the sun sweep equal area in equal time (Kepler’s second law). So we see that conservation of angular momentum is equivalent to Kepler’s second law. Solving (b) for dθ/dt and substituting the result into (a), we have 2  d2 r L2 L d2 r = −f (r) , or equivalently m = − f (r) . m 2 − mr dt mr2 dt2 mr3 Newton’s universal law of gravitation postulates that GM m f (r) = , r2 where M and m are the masses of the two bodies, G is a universal physical constant, and r is the distance between the two objects. So the equation we must solve is L2 GM m d2 r = − . (c) dt2 mr3 r2 The trick to solving this is to make the substitution, r = 1/u. Then differentiating r once yields 2 u 1 du 1 du dθ 1 du L dr = −m 2 = −m 2 = − m 2 . m dt u dt u dθ dt m u dθ   m

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Differentiating a second time yields d2 r d du d2 u dθ L2 u2 d2 u = −L . = −L = − dt2 dt dθ dθ2 dt m dθ2 Substituting this result into (c) we get m

L2 u2 d2 u L2 u3 = − GM mu2 , or equivalently m dθ2 m The general solution to (d) (see case 3 on page 309) is −

u = u0 cos (θ + φ) +

GM m2 ; L2

back-substuting, r =

GM m2 d2 u + u = . dθ2 L2

(d)

L2 , (e) u0 L2 cos (θ + φ) + GM m2

where u0 and φ are arbitrary constants. Equation (e) does not tell us anything about the time-behavior of the orbit, but it does give the shape of the orbit. Setting φ = 0 aligns the symmetry of the orbit with the x-axis. By substituting r2 = x2 + y 2 and x = r cos θ, we find that the resulting Cartesian formula is the equation of a conic section, of which an ellipse with one of the foci at the origin is one of the possibilities fitting the equation. The latter is mathematical confirmation of Kepler’s first law. Replacing φ with nonzero values simply rotates the major axis of the ellipse away from the x-axis. The time-behavior of the orbit can only be determined by numeric methods. Such methods, as well as more details on the derivation above, can be found in An Introduction to Celestial Mechanics by Forest Ray Moulton (1914 MacMillan. Available since 1970 from Dover Publications). Introduction to Fourier’s Heat Equation and its Solution: This topic is covered here only as the barest of outlines. Imagine a long thin strand of heat conducting material, such as a length of wire. We also imagine that it is completely insulated from the outside world except maybe at its endpoints. So heat conduction is restricted to heat traveling along the length of the wire. A flow of heat along the wire results in a temperature gradient according to the constant, σ. That is, for example, if σ = 1 calorie/C◦ cm second, then if the temperature gradient is 1 C◦ per cm, heat flow past any point of that gradient will be 1 calorie per second. We also let the thermal density of the wire be represented by ρ. Then, for example, if ρ = 1 calorie/cm C◦ and a point on the wire is at temperature 100 C◦ , then that point on the wire contains 100 calories/cm of heat. If we take the case where one end of the wire of length, L, is held at temperature, T1 and the other at T2 , and the temperature gradient is constant throughout the length at a rate of (T2 − T1 ) /L, then heat flow will be σ (T2 − T1 ) /L. If the initial heat gradient is constant at this value, then as long as the ends are held at T1 and T2 , the heat gradient along the entire length of the wire will remain unchanged over time. That is, a constant heat gradient of (T2 − T1 ) /L is the steady state condition for the boundary conditions of T1 and T2 held at the respective ends of the wire. Having constant heat gradient implies that ∂ 2 T /∂x2 = 0, where x represents position along the wire, and whenever this condition is true, the wire is in steady state, which is to say that the temperature profile will remain unchanged through time. But when ∂ 2 T /∂x2 6= 0, we find that the temperature profile changes over time even if the boundary conditions at either end of the wire are held constant. Finding the time function of the temperature profile under this condition is the problem Jean Baptist Joseph Fourier (1768-1830) sought to solve in 1822 (see God Created the Integers, a collection of mathematical breakthrough papers edited and commented by Stephen Hawking, Running Press, 2005 for more on Fourier and his work).

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We can see intuitively that wherever there is a “hump” in temperature, that is where ∂ 2 T /∂x2 < 0, there is a net outward flow of heat from that region to dissipate the hump, and wherever there is a “dip” in temperature, that is where ∂ 2 T /∂x2 > 0, there is a net inward flow of heat to that region to “fill up” the dip. Fourier’s heat equation, 1 ∂T ∂2T = , (a) ∂x2 k ∂t where k = σ/ρ quantifies this. This equation is similar to example 1 on page 351, and page S-52 offers a solution to this equation in terms of decaying sinusoids, X 2 T (x, t) = Cω e−kω t sin (ωx + φω ) , ω

or equivalently

T (x, t) =

X

e−kω

ω

2

t

{Aω cos (ωx) + Bω sin (ωx)} ,

(b)

where Aω and Bω are sets of constants, one A and one B for each value of ω used in the solution. What Fourier recognized and is remembered for is the observation that you can always convert any initial bounded function, T (x, 0), into a weighted sum of sines and cosines, and likewise, you can always convert any weighted sum of sines and cosines into a function of x. So the procedure is to first convert the initial T (x, 0) into a set of Aω ’s and Bω ’s for some set of values of ω. Second, to find what has happened after 2 some elapsed time, t, we apply the decay function, e−kω t , for each value of ω to the corresponding sines and cosines. Third and finally, we sum up the decayed sines and cosines each weighted by the Aω ’s and Bω ’s that we found in the first step. If we restrict our consideration of a finite length, L, of the wire, then the set of ω’s we need to consider is ωk = 2πk/L, for each nonnegative integer, k. To demonstrate how we break an arbitrary function T (x, 0), defined over 0 ≤ x ≤ L, into sines and cosines, we first demonstrate the principle of orthogonality of sines and cosines. Observe that Z L cos (ωj x) sin (ωk x) dx = 0 for any j and k 0

and Z L

cos (ωj x) cos (ωk x) dx = 0 and

Z

L

sin (ωj x) sin (ωk x) dx = 0

0

0

whenever j 6= k

In words, we have a collection of functions, the product of any two of them when integrated from 0 to L yields zero, provided the two functions are distinct. Such functions are said to be orthogonal to one another under this operation. The operation itself is fundamental to this problem and we give it the name, scalar product. We define the scalar product of two function, f (x) and g (x) as Z L f (x) g (x) dx, hf, gi = 0

and we say that nonzero f and g are orthogonal whenever hf, gi = 0. If we assume that the function, T (x, 0), is the sum, X T (x, 0) = Ak cos (ωk x) + Bk sin (ωk x) , k

for some set of Ak and Bk , then taking the scalar product of T with any any cos (ωj x) yields hT (x, 0) , cos (ωj x)i = hAj cos (ωj x) , cos (ωj x)i . Observe that by orthogonality, all the terms in the original sum except the one above to the right of the equal are zero. Only for the cosine term where k = j do we get a

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nonzero term. Likewise with hT (x, 0) , sin (ωj x)i = hBj sin (ωj x) , sin (ωj x)i .

It is easily verified using our definition of scalar product that hAf (x) , g (x)i = A hf (x) , g (x)i

and

hf (x) , Bg (x)i = B hf (x) , g (x)i ,

for any functions, f and g, and for any constants, A, and B. This makes it easy to solve for the value of any Ak or Bk . Simply take the product of T (x, 0) with corresponding cos (ωk x) or sin (ωk x) and integrate the product from 0 to L, then divide by L/2. The dividing by L/2 is because L for all k. hcos (ωk x) , cos (ωk x)i = hsin (ωk x) , sin (ωk x)i = 2 Proving this last fact is left as an exercise for the reader. To prove that the desired set of Ak ’s and Bk ’s always exists, we sample the interval, [0, L] at 2n + 1 equally spaced points. iL for i from 0 to 2n. xi = 2n We also isolate 2n + 1 sinusoid functions, cos (ωk x) for k from 0 to n and sin (ωk x) for k from 1 to n. Note that we omit consideration of sin (ω0 x) because it is zero for all x. We allow rows k = 0 through n of a matrix, M , to be cos (ωk xi ), where i goes from 0 to 2n. We allow rows k = n + 1 through 2n to be sin (ωk−n xi ). This gives us an invertible square matrix. Then taking a vector of n + 1 Ak ’s and n Bk ’s times the matrix, M , gives a weighted sum of the sine and cosine functions over the various sample points, xi . It follows that M −1 T (xi , 0) must give back a vector of the weights, Ak and Bk . As we let n go to infinity, we see that we can reproduce the original function, T (x, 0), on a denser and denser set of sample points. There is much more can be said about solving Fourier’s heat equation. Indeed there are entire textbooks devoted to this one equation. The point here is to introduce you to the notion of decomposing a function into weighted sinusoids. This procedure is known as Fourier analysis, and is the topic of chapter VIII of J.W. Mellor’s book. It is also applicable to a wide variety of problems other than the solving Fourier’s heat equation. In an earlier section, for example, where we made Bode plots of the responses of resonant systems to sinusoidal inputs, we were laying the groundwork for applying Fourier analysis to this problem as well. In real life the inputs to such systems are not sinusoids, but could be any function. But the fact that any legitimate input function can be decomposed into sinusoids allows us to solve the action of the system on each sinusoid, then sum the output sinusoids to determine the system’s response to the original input. Even the action of the human ear is connected with Fourier’s method. The human ear is an instrument that resolves an input function, namely sound acting upon the eardrum, into a set of sinusoids of various frequencies and transmits the magnitudes of the resolved frequencies to the brain. Also be aware that Fourier analysis is just a single example of a more general methodology of decomposing an arbitrary function into a weighted sum of some set of basis functions. In the more general setting, the basis set need not be sinusoids. Depending upon the type of problem, resolution into other basis functions is often convenient. As mentioned on page 358 of the text, Bessel functions provide a useful basis for systems that have cylindrical symmetry. And there are others for other symmetries.

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Selected referenced sections of original text begin

§ 20. Leibnitz’ Theorem. To find the nth differential coefficient of the product of two funnctions of x in terms of the differential coefficients of each function. On page 26 the differential coefficient of the product of two variables was shown to be d (uv) du dv dy = =v +u , dx dx dx dx where u and v are fuuctions of x. By successive differentiation and analogy with the binomial theorem it may be shown that dn u dv dn−1 u dn v dn (uv) =v n +n + ··· + u n. (1) n n−1 dx dx dx dx dx This formula, due to Leibnitz, will be found very convenient in Chapter VII “How to Solve Differential Equations”. The reader must himself prove the formula, as an exercise on (§ 19, by comparing the values of d2 (uv)/dx2 , d3 (uv)/dz 3 ,..., with the developments 2 3 of (x + h) , (x + h) ,..., of page 22.

49

The coefficient, n, is a binomial coefficient, as are the respective unshown coefficients in this equation. KH

§ 22. Euler’s Theorem on Homogeneous Functions. The following discussion is convenient for reference: To show that if u is an homogeneous function∗ of the nth degree, say u = Σaxα y β ,†where α + β = n, then ∂u ∂u x +y = nu. (14) ∂x ∂y By differentiation of the homogeneous function, P 1 1 u = axα y β + bxα y β + · · · = axα y β ,

where α + β = α1 + β 1 = · · · = n, we obtain ∂u P ∂u P = aαxα−1 y β and = aβxα y β−1 . ∂x ∂y Hence P ∂u ∂u P x +y = a (α + β) xα y β = n axα y β . ∂x ∂y The theorem may be extended to include any number of variables (see footnote, page 340). ∂u ∂u EXAMPLES.– (1) If u = x2 y + y 2 x + 3xyz, then x ∂u ∂x + y ∂y + z ∂z = 3u. Prove this result by actual differentiation. It of course follows directly from Euler’s theorem, since the equation. is homogeneous and of the third degree. 3 2 y+y 3 ∂u ∂u (2) If u = xx2+x +xy+y 2 , x ∂x + y ∂y = u, since the equation is of the first degree and homogeneous. (3) Put Euler’s theorem into words. Ansr. In any homogeneous function, the terms of the products of each variable with the partial differential coefficients of the original function with respect to that variable is equal to the product of the original function with its degree. ∗ An homogeneous function is one in which all the terms containing the variables have the same degree. Examples: x2 + bxy + z 2 ; x4 + xyz 2 + x3 y + x2 + z 2 are homogeneous functions of the second and fourth degrees respectively. †The sign “Σ” is to be read “the sum of all terms of the same type as. . . ” or here the sum of all terms containing x, y and a constant”. The symbol “Π” is sometimes used in the same way for “the product of all terms of the type”.

56

§ 23. Successive Partial Differentiation. We can get the higher partial derivatives by successive differntiation, using processes analogous to those used on page 47. Thus when u = x2 + y 2 + x2 y 2 , ∂u = 2x + 2y 3x; ∂x repeating the differentiation,
∂2u 3 = 2 1 + y ; ∂x2

∂u = 2y + 3x2 y 2; ∂y
∂2u 2 = 2 1 + 3x y ; ∂y 2

(1)

(2)

If we had differentiated ∂u/∂x with respect to y, and ∂u/∂y with respect to x, we should have obtained two identical results, viz.:–

∂2u ∂2u 2 = 6y x, and = 6y 2 x. (3) ∂y∂x ∂x∂y This rule is general. The higher partial derivatives are independent of the order of differentiation. By differentiation of ∂u/∂x with respect to y, assuming x to   ∂2u ∂u ∂y, which is written ; on the other be constant, we get ∂ ∂x ∂y∂x hand, by the differentiation of ∂u/∂y with respect to x, assuming y to be ∂2u constant, we obtain . That is to say ∂x∂y ∂2u ∂2u = . ∂y∂x ∂x∂y

(4)

This was only proved in (3) for a special case. As soon as the reader. has got familiar with the idea of differentiation, he will no doubt be able to deduce the general proof for himself, although it is given in the regular text books. The result stated in (4) is of great importance. § 24. Exact Differentials. To find the condition that u may be a function of x and y in the equation du = M dx + N dy, where M and N are functions of x and y. We have just seen that if u is a function of x and y ∂u ∂u du = dx + dy, ∂x ∂y that is to say, by comparing (5) and (6) M=

∂u ∂u ; N= . ∂x ∂y 57

(5)

(6)

Differentiating the first with respect to y, and the second with respect to x, we have, from (4) ∂M ∂N = . (7) ∂y ∂x In the chapter on differential equations this condition is shown to be necessary and sufficient in order that certain equations may be solved, or “integrated” as it is called. Equation (7) is therefore called the criterion of integrability. An equation that satisfies this condition is said to be a complete or an exact differential. For examples, see page 290. § 25. Integrating Factors. The equation M dx + N dy = 0

(8)

can always be made exact by multiplying through with some function of x, called an integrating factor. (M and N are functions of x and y.) Since M and N are functions of x and y, (8) may be written M dy =− (9) dx N. or the variation of y with respect to x is as −M is to N; that is to say, x is some function of y, say f (x, y) = a, then from (5), page 52, ∂f (x, y) ∂f (x, y) dx + dy = −. ∂x ∂y

(10)

By a transformation of (10), and a comparison of the result with (9), we  find that dy ∂f (x, y) ∂f (x, y) M =− dx dy = − . (11) dx ∂x ∂y N ∂f (x, y) ∂f (x, y) = µ M; and = µ N, (12) Hence ∂x ∂y where µ, is either a function of x and y, or else a constant. Multiplying the original equation by the integrating factor µ, and substituting the values of µ M, µ, N so obtained in (12), we obtain ∂f (x, y) ∂f (x, y) dx + dy = 0, ∂x ∂y which fulfils the condition of exactness. EXAMPLE.– Show that the equation y x dy = 0 becomes exact when multiplied by 1/y 2 . ∂M ∂N 1 1 = − 2; = − 2. ∂y y ∂x y Hence ∂M/∂y = ∂N/∂x, the condition required by (7). In the same way show that 1/xy and 1/x2 are also integrating factors.

58

§ 67. Envelopes. m The equation + ax, a represents a family of curves, since for each value of a, we get a distinct curve. If a varies continuously it will determine a succession of curves, each of which is a member of the family denoted by the above equation. a is said to be the variable parameter of the family, since the different members of the family are obtained by assigning arbitrary values for a. Let the equations m + ax (1) y1 = a m + (a + δa) x (2) y2 = a + δa m y3 = + (a + 2δa) x (3) a + 2δa 142

be three successive members of the family. As a general rule two distinct curves in the same family will have a point of intersection. Let P (Fig. 78) be the point of intersection of curves (1) and (2); P1 the point of intersection of curves (2) and (3), then, since P1 , and P2 , are both situated on the curve (2), P P1 , is part of the locus of a curve whose are P P1 , coincides with an equal part of the curve (2). It can be proved, in fact, that the curve P P ,. . . touches the whole family of curves represented by the original equation. Such a curve is said to be an envelope of the family. To find the equation to the envelope, bring all the terms of the original equation to one side, Fig. 78. – Envelope m y= − ax = 0. a Then differentiate with respect to the variable parameter, and put m − x = 0. a2 Eliminate a, between these equations, r √ √ m y− m·x−x = 0, or, y − 2 m · x = 0. x ∴ y 2 = 4mx. EXAMPLES.– Find the envelope of the family of circles 2

(x − a) + y 2 = r2 ,

where a is the variable parameter.

Differentiate with respect a and x − a = 0;

eliminating a, we get y = ±a, which is the required envelope. The envelope y = ±a represents two straight lines parallel to the x-axis and at a distance +a, and −a from it. Shown Fig. 79. (2) Show that the envelope of the family of curves (x − m − a)2 + y 2 = 4ma, is a parabola y 2 = 4mx. See §§ 126 and 138. Fig. 79. – Double Envelope

143

§ 72. How to find a Value for the Integration Constant. It is perhaps unnecessary to remind the reader that integration constants must not be confused with the constants belonging to the original equation. For instance, in the law of descent of a falling body R R dv/dt = g; dv = g dt, or, v = gt + C. (1)

Here g is a constant representing the increase of velocity due to the earth’s attraction, C is the constant of integration. The student will find some instructive remarks in § 118. There are two methods in general use for the evaluation of the integration constant. First Method. Returning to the falling body and to its equation of motion, v = gt + C.

On attempting to apply this equation to an actual experiment, we should find that, at the moment we began to calculate the velocity, the body might be moving upwards or downwards, or starting from a position of rest. All these possibilities are included in the integration constant C. Let v0 , denote the initial velocity of the body. The computation begins when t = 0, hence v0 = g × 0 + C,

or, C = v0 .

If the body starts to fall from a position of rest, v0 = C = 0, and R dv = gt, or, v = gt.

This suggests a method for evaluating the constant whenever ’the nature of the problem permits us to deduce the value of the function for particular values of the variable. If possible, therefore, substitute particular values of the variables in the equation containing the integration constant and solve She resulting expression for C. EXAMPLE.– Find the value of C in the equation t=

1 1 log + C, k a−x

(2)

which is a standard “velocity equation” of physical chemistry. t represents the time required for the formation of an amount of substance x. When the reaction is just beginning, x = 0 and t = 0. Substitute these values of x and t in (2). 1 1 1 1 log + C = 0, or, C = − log . k a k a Substitute this value of C in the given equation and we get   1 1 1 a 1 log = log − log . t= k a−x a k a−x

Second Method. Another way is to find the values of x corresponding to two different values of t. Substitute the two 162

sets of results in the given equation. The constant can then be made to disappear by subtraction. EXAMPLE.– In the above equation, (2), assume that when t = t1 , x = x1 and when t = t2 , x = x2 ; where x1 , x2 , t1 , and t2 , are numerical measurements. Substitute these results in (2). 1 1 1 1 t1 = log + C; t2 = log + C. k a − x1 k a − x2 By subtraction and rearrangement of terms a − x1 1 . t2 − t1 = log k a − x2 The result of this method is to eliminate, not evaluate the constant.

Numerous examples of both methods will occur in the course of this work. Some have already been given in the discussion on the “Compound Interest Law in Nature”.

163