Rudolph E. Langer a First Course in Ordinary Differential Equations

A First Course in

ORDINARY

DIFFERENTIAL

EQUATIONS

A First

Course in

ORDINARY DIFFERENTIAL EQUATIONS

Rudolph E. Longer, Ph.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF WISCONSIN

N E W Y O R K J O H N W I L E Y & S O N S , I N C L O N D O N C H A P M A N & HALL , L I M I T E D

Copyright, 1954, By John Wiley & Sons, Inc.

All Rights Reserved This book or any part thereof must not be reproduced in any form without the written permission of the publisher.

Library of Congress Catalog Card Number: 54-5326

Printed In the United States of America

PREFACE

In presenting this textbook I have indulged a long-felt desire to teach on a wider range a subject that has been my main interest over many years. Differential equations are so important in their applications that no scien-tist and no engineer can afford to neglect them. A t the same time they are the carriers of so many fundamental, far-reaching, and ingenious ideas that mathematical analysis can well be regarded as centered around them. Present-day research upon them is alive and intensive, and is being spurred continuously by the urgent needs of the most up-to-date developments in science and technology.

This is a first course i n the subject, written for the student with a year's experience in the calculus. Its manipulative material is adjusted to that level, and whatever goes beyond it is explained in detail. In connection with each new development, examples of al l the various cases that may arise are given and are worked out at every step. These wi l l serve as guides of method, and also as models of economical procedure. T h e reader may miss a chapter on review material. Such material, in con-siderable amount, is nevertheless there, generally in the explanatory examples. It has always seemed to me preferable on pedagogical grounds to insinuate the reviews unobtrusively where the motivations for them exist, rather than to emphasize his shortcomings to the student by fencing off the main subject by a discouraging barrier of material that may be relevant here or there in the book.

In the matter of method, I have consistently played down the need for memorization and played up the common sense and reasonable approach which experience leads one to recognize. N o tabulations of formulas or lists of rules are therefore to be found in the book. Although I have not concealed any essential difficulties (my intention is that the book should be wholly honest) I have tried consistently to disburden the presentation of unnecessary difficulties. T o that end I have avoided such things as untimely appeals to existence theorems when explicit solutions are avail-able, and where proofs of substantial length have been included I have

v

vi Preface

stripped them of generalities not germane to the present context to reduce them to their most elementary and perspicuous terms. I have kept close to the subject. Although the applications to physics, chemistry, engineer-ing, etc., have been stressed, this is still a book on differential equations, not a potpourri of textbooks on these other subjects.

I believe that all the material of the book may be taken up profitably, and in the order i n which it is here presented, where time permits. How-ever, continuity wi l l not be sacrificed even by many and substantial omissions. A shorter course may well be based upon Chapters 1, 2, and 3; parts of Chapters 4 and 5; and Chapters 8 and 9. I have assumed that the student wi l l have a table of integrals at hand.

Although I have broached the significance of the differential equation, the arbitrary constant in its solution, and such matters, even in the intro-ductory Chapter 1, it is my belief that some substantial first-hand experi-ence with the equations alone supplies a basis upon which more extended discussions of these matters is easily comprehensible and truly meaningful. I have accordingly, in Chapters 2 and 3, hastened into the manipulative work of finding explicit solutions and applying them. This wi l l engage the student almost at once in operations which are already familiar to h im from the calculus, and wi l l give him the opportunity to establish confidence in his ability to do something definite. A t the same time it wi l l make the essential characteristics of the differential equation concretely familiar to hira.

In Chapter 2, I have presented the explicit methods that are applicable to differential equations of the first order in a definite relationship to the two fundamental notions of the integrating factor and the change of varia-bles. This gives organization to material that has all too often been presented as hardly more than a collection of unrelated tricks. In Chapter 4 the geometrical significance of the differential equation, as it is displayed by the integral curves, is stressed. Chapter 5 centers upon the procedures of approximation to be resorted to when explicit solutions cannot be found. I have included at this point a proof of the existence and uniqueness of a solution under appropriate conditions, in the belief that this reasoning wi l l be manageable and challenging to the more able student. For others it may be omitted. Chapters 8 and 9, on the differential equation of the second order with constant coefficients, wi l l be of prime importance to students of physics and engineering. In Chapter 11,1 have set forth a more complete treatment of solutions by power series than is usual in books of this kind.

It has been my steady purpose throughout the book to present the dif-ferential equation dually as a mathematical concept and as a technological tool. Consistent with this I have tried to convey an understanding of

Preface

what such an equation is and involves, and no less to give practice in the commoner techniques that are available for solving it , or otherwise extracting useful information from it.

R U D O L P H E . L A N G E R

Madison, Wisconsin January, 1954

CONTENTS

Chapter Page 1. I N T R O D U C T I O N 1

1.1. What is a differential equation? 1 1.2. Integrals or solutions of a differential equation 2 1.3. The direction field of a differential equation 3 1.4. Constants of integration 5 1.5. Checking a general solution. The differential equation of

a family of curves 5

2. S O M E T Y P E S O F S O L V A B L E D I F F E R E N T I A L E Q U A -T I O N S O F T H E F I R S T O R D E R 9

2.1. Equations with separable variables 9 2.2. Exact differential equations 11 2.3. Integrating factors 14 2.4. Finding integrating factors 15 2.5. The linear differential equation of the first order 17 2.6. Changes of variables 19 2.7. Equations with homogeneous coefficients 22 2.8. Equations with linear coefficients 24 2.9. Simultaneous equations 26 2.10. The linear differential system with constant coefficients 28 2.11. The replacement of a single differential equation by a dif-

ferential system 29

3. A P P L I C A T I O N S O F D I F F E R E N T I A L E Q U A T I O N S O F T H E F I R S T O R D E R 33

3.1. Problems in velocities 33 3.2. Frictional motion 37 3.3. Problems in rates 38 3.4. Problems of flow 40 3.5. Locus problems. Rectangular coordinates 44

ix

X Contents

Chapter Page

3.6. Polar coordinates 47 3.7. Some curves of pursuit 49 3.8. Suspension cables 54 3.9. The rope around a shaft 58

4. I N T E G R A L C U R V E S . T R A J E C T O R I E S 63 4.1. Integrals of a differential equation 63 4.2. The direction field of a differential system 64 4.3. The isoclines of a direction field 65 4.4. Traj ectories 68 4.5. Singular points of a direction field 70 4.6. Some elementary cases of the differential equation

' _

^ ~ hx + ky 72

4.7. The integral curves as trajectories 73

5. A P P R O X I M A T E S O L U T I O N S . I N F I N I T E S E R I E S . E X I S T E N C E A N D U N I Q U E N E S S O F A S O L U T I O N 78

5.1. Polygonal graph approximations 78 5.2. Polygonal approximations for simultaneous equations 81 5.3. Solutions in power series. Polynomial approximations 83 5.4. The method of successive approximation 88 5.5. Successive approximations for differential systems 91 5.6. A proof of the existence of a solution 93 5.7. Uniqueness of the solution 96

6. F U R T H E R S T U D I E S O F D I F F E R E N T I A L E Q U A T I O N S O F T H E F I R S T O R D E R . T H E R I C C A T I E Q U A T I O N 98

6.1. Singular solutions 98 6.2. Factorable equations 100 6.3. Singular solutions again 101 6.4. Integrations by differentiation. The Clairaut equation 103 6.5. The method of difTcrcntiation with respect to j 105 6.6. The Riccati equation 107 6.7. Some properties of the integrals of Riceali equations 110

7. S O M E S O L V A B L E D I F F E R E N T I A L E Q U A T I O N S O F T H E S E C O N D O R D E R . A P P R O X I M A T I O N S . AP-P L I C A T I O N S 114

7.1. Differential equations of the second order 114

Contents xi

Chapter Page

7.2. Equations in which either or x are not present 115 7.3. The differential equation of a 2-parameter family of curves.

Checking a general solution 116 7.4. Exact equations 118 7.5. Simultaneous equations 120 7.6. Solutions i n power series. Polynomial approximations 122 1.1. Approximate integral curves 123 7.8. Some geometrical applications 127 7.9. Curves of pursuit 128 7.10. The motion of a particle on a curved path 130

8. T H E L I N E A R D I F F E R E N T I A L E Q U A T I O N W I T H C O N -S T A N T C O E F F I C I E N T S 134

8.1. The complete equation 134 8.2. The complementary function 135 8.3. T h e case of complex roots. Euler's formulas 137 8.4. Differential operators 140 8.5. A n operational method for finding a particular integral 141 8.6. A second operational method 144 8.7. A comparison of formulas 146 8.8. The method of undetermined coefficients 147 8.9. The method continued 149 8.10. Difierential equations of higher order than the second 152

9. A P P L I C A T I O N S O F L I N E A R D I F F E R E N T I A L E Q U A -T I O N S O F T H E S E C O N D O R D E R 156

9.1. Simple harmonic motion 156 9.2. Free vibrations 159 9.3. Damped vibration 163 9.4. Forced vibration without damping. Resonance 166 9.5. Forced vibration with damping 169 9.6. The forced vibration of a system with a repelling force 172 9.7. Mot ion under an intermittent force 173 9.8. Simple electrical circuits 176

180 180

10. T H E L I N E A R E Q U A T I O N W I T H V A R I A B L E CO-E F F I C I E N T S

10.1. Ordinary and singular points 10.2. The Euler equation 181 10.3. Exact equations 182 10.4. The adjoint differential equation. Integrating factors 183

xii Contents

Chapter Page

10.5. The existence and uniqueness of a solution. The Riccati equation 185

10.6. The Wronskian. Linear independence of solutions 186 10.7. The general integral 188 10.8. The change of dependent variable 189 10.9. Solution of an equation when an integral of the reduced

equation is known 190 10.10. The method of variation of parameters 193 10.11. The change of independent variable 194 10.12. Simultaneous linear equations 196 10.13. The Laplace transformation 197

11. S O L U T I O N S I N P O W E R S E R I E S 201 11.1. The reduced equation at an ordinary point 201 11.2. A proof of convergence 203 11.3. The complete equation at an ordinary point 205 11.4. T h e ordinary point at . Solutions in powers of Xjx 206 11.5. The reduced equation at a regular singular point 208 11.6. Logarithmic integrals of the reduced equation 211 11.7. The complete equation 213 11.8. The regular singular point at

CHAPTER

1

Introduction

1.1. What is a differential equation?

The central problem of the integral calculus is this: From a given func-tion f(x)y to find an integral

and to apply this result to the solution of a problem. In terms of symbols this may be stated as follows: Given the equation (1.1) dy/dx=f{x\

to find functions >(x) which fulfill it , and to apply them. It is but a short step from an equation 1.1 to one of the more general

form

(1.2) F{x,y,y') = 0 ,

in which y' stands for dy/dx. A n y equation of form 1.2 that actually involves^' is called an ordinary differential equation of the first order. A n equa-tion that involves the second derivative oiy but no derivative of a higher order is called an ordinary differential equation of the second ordery and one that involves the nth derivative, but none of higher order, is called a differential equation of the nth order. The term "ord inary" refers to the fact that the derivatives are ordinary ones (not partial derivatives).

The reader wi l l already know that equations of type 1.1 have a wide range of scientific and technological applications. The more inclusive type of equation 1.2 is even much more widely applicable. Indeed, the differential equation is the primary mathematical instrument for the pre-cise expression of nature's laws, for these laws are always statements of relationships between variables and their rates of relative change, namely, their derivatives. For that reason the development of the theory of differ-ential equations has been hastened from the time of its very beginning by the need for solutions of scientific problems. However, this theory also has great interest for reasons quite apart from its applications. It is a

1

2 Introduction

branch of mathematical analysis that has occupied the minds of many of the great mathematical masters, and continues, even i n the present, to be an active field of mathematical research.

1.2. Integrals or solutions of a differential equation A n y function that fulfills a differential equation is called an integral^ or a

solution of that equation. T o solve a differential equation is to make these functions known. This statement is precise only if the sense of the term " k n o w n " is made clear. A function is "known, " of course, if it can be expressed by a formula in terms of standard and familiar functions. We shall, however, use the term in a broader sense by admitting a number of alternatives. In each of these the values of the function can be found by calculation, at least to any prescribed degree of accuracy.

(i) If a function f{x) is known, and is integrable on an interval contain-ing the point x = a, the function

is regarded as "known. " We shall henceforth call this function a

The Direction Field 3

Since; ' ' = dy/dx, such an equation is often conveniently put into the form

(1.4) P{x, y) dx ^ Q{x,y) dy = 0.

This is accomplished, of course, by muhiplying the equation by Q{xj y) dx, and then writing P{x,y) in place of ~(l{xyy)f{xyy). Equations 1.4 and 1.3 are, however, not necessarily equivalent. Equation 1.4 may, for instance, be fulfilled by the relation x = c. However, x = c, since it does not characterize; ' , has no significance for equation 1.3. Also, equation 1.4 may be fulfilled by the relation Cl{x,y) = 0, without equation 1.3 being so.

The choice of letters to designate the variables in a differential equation is immaterial. Thus we may have an equation

F{t, s, ds/dt) = 0,

whose integrals are s = f{t) or ) has a numerical value. There is a direction in the plane whose slope has that vsdue. By assigning this direction to the point, and doing so for each point, we impose upon the region a so-called direction field. Clearly every differential equation 1.3 defines a direction field in any region i n which its function f{x, y) has the properties specified above. A direction field may be represented graphically, though somewhat crudely, by drawing short line segments in the prescribed directions through a number of its points. Such a representation is shown in F ig . 1, which applies to the differential equation

The function fix^y) is in this case continuous except on the line x = 2. The slope of this field is at any point twice that of the line joining that point with the point (2, 1).

Direction fields are common in our physical surroundings. The force of gravity surrounds the earth with such a field, and its magnetism likewise

4 Introduction

does so. A n y flow of a l iquid or gaa defines such a field at each instant, the direction of the field being that of flow.

In the instance of any direction field we may conceive of a particle as starting from any one of its points, and then moving always i n the direction of the field. The path of such a particle is a streamline of the field. E v i -

y

/

FIG. 1.

dently every point of a field lies upon a streamline. Now a streamline is a graph. It defines > as a function of x. A t any of its points its slope i s^ ' . It is, however, also /(x, namely, the slope of the field. W e see thus that the function^ defined by a streamline is an integral of the differential equation to which the field applies. The differential equation thus has a

FIG. 2.

whole family of integrals corresponding to the family of streamlines Figure 2 shows some streamlines for the differential equation

y = W ~ X- +^x - iy.

The streamlines are called the integral curves of the differential equation.

Checking a Generell Solution 5

1.4. Constants of integration A differential equation of the simple type 1.1 is solved by a straight-out

quadrature. This introduces a constant of integration, namely, a con-stant which is arbitrary in the sense that it may be given any value on a suitable range. There is therefore a family of integrals. It is a 1-parame-ter family, because just 1 constant identifies its several members. It appears that any equation 1.3 also has a 1-parameter family of integrals, for, if we think of its streamlines as cut by some transversal curve, each point of that curve is determined by a single parameter value, for instance, by its distance along the curve from some specified point. Each point also identifies the streamline through it. Thus the integral curves are identifi-able by the values of 1 parameter. In the coordinate equation for the family an arbitrU"y constant therefore appears,

A relation that solves a differential equation 1.2 and that contains an arbitrary constant is called the general integral, or the general solution. A n integral that is obtainable by giving to the constant a specific value is called a particular integral. The general integral may, and often does, include al l the integrals of the differential equation. There are, however, cases in which it does not do so. A n integral that is not so included is called a singular integral.

Example 1. The differential equation

y" + 2{l -y] = 0

has the general integral

However, it also has_y = 1 as an integral, y = ^, since it is not included in the general integral, is a singular solution.

1.5. Checking a general solution. The differential equation of a family of curves

T o check a relation

(1.5)

6 Introduction

yy and y'. If it is the same as the original differential equation or the equivalent of it , the integral is thereby checked. If equation 1.6 does not contain f it is itself the liminant.

Example 2. Given the differcntiad equation

* + ^ | / + > - 2 x = 0.

T o check the relation xy \o% y x"^ ~ c

as its general integral. The derivative of the given relation is

x y ' + > + - - 2 x = 0. y

This is the original differential equation. The integral has thus been checked.

Example 3. Given the differential equation

^ 1 T o check the relation

y = log c tan ' x\

as its general integral. The derivative of the given relation is

- 1 y = {1 + x 2 ) { c - tan-^j}

The liminant of c is thus v' = This is the original dif-{1 + x-\'

ferential equation. The integral is thus checked.

The method described may be used to find the differential equation of any 1-parameter family of curves. Thus, let the family have the coordi-nate equation 1.5. By differentiation we may obtain equation 1.6. The liminant of c is the differential equation of the family.

Example 4. T o find the differential equation of the family of curves

log> +

Checking a General Solution 7

The derivative of this equation is

- + 2cx = 0 y

The eUminant of c is the relation

logjy - J - 1 = 0

This is the required differential equation.

ASSIGNMENTS In each of the following cases, check the relation given as the general integral of the

respective differential equation.

1. =

2. y = cx^,

3. y = c cos X + sin x,

4. 7 = + c tan X,

5. ^ + log ^ f = 0,

6. sin^ ^ + 2 tan x =

for yy' + X = 0.

for xy' -2y = 0.

for y' cos x + ^ sin x 1 0 .

for \y' ^] sin X {7 } sec x 0.

for y' -{-ytr" = 0.

for y' +2 scc^ y esc i2y) 0.

7. y - 2 1 + cx

8. \x+cU--\

= 0. for y' y - 2

= 0,

for y' +2xy = ''\

9. y{x + 4| + + = 0, for {x + 4} / + {x + 5]y - 2e^ = 0.

10. = 1

2 cos X + c sin X

11. log (xy) = cx.

for y' -\- -y cot X ^ ' CSC X = 0.

for xy' +y = y log {xyj.

2 . sin * X + sin * ^ =* - 1 for + 1 V i ' V i -

- = sm * X + sm ^ y.

PROBLEMS In each of the following cases, find the differential equation for the given family of

curves.

i. y cx ^ x'.

3. y 2 ^ c tan x,

5. + Ic^ X = 0,

2, 33 J

2 I i I 112 2 4. U + ^ r + [;r + l |

6. + 1,

1 8, y. cx 1.

8 Intrcxiuction

9. |c + > P + 2:Jt = 0. 10. -y/y -X + log \cx\ - 1

11- > " i T T - 12. - Jf + 1 * 2 cos + c sin X

ANSWERS TO ODD-NUMBERED PROBLEMS

\. xy' ~y ~2x* 3. / - ^ ~ ^ sm X cos X

5. log X - ^ - 0. 7. ' ' 1 + X - i - V 1 +

9. ^ - 2xy ' ( / + 1 1 - 0 .

11. 2(x^ - X + + {2x - 1 1 / - 2x + 1.

CHAPTER

2

Some Types of Solvable Differential Equations of the First Order

2.1. Equations with separable variables

The central problem with which we shall be concerned is that of study-ing the integrals of given differential equations. We may attempt to find these integrals exactly or approximately, and we may do either of these things analytically by the use of methods of the calculus, or by the use of graphs, direction fields, etc. As we shall see, only certain types of differ-ential equations can be solved explicitly. In this chapter some such types are considered. Their integrals are applicable to many practical prob-lems. They are also typical of the integrals of many differential equations that cannot be explicitly solved.

A differential equation

(2.1) P{x, y) dx + Q{x, y) dy ^ 0,

i n which P(x, y) and Q(x, y) are each a product of a function of x by a func-tion of 7, is said to have separable variables. We may write such an equation more explicitly as

Pi{x)p2{y) dx + qi[x)q2{y) dy = 0.

If there is a value yo at which ptiyo) ~ 0, the equation is fulfilled by the relation y >o since that gives dy 0. Thus ^ = is a solution. If there is a value ;ifo at which q\{xQ) = 0, the equation is also fulfilled by the relation x = xo. This , however, does not define ^ as a function of x.

For values of x and^ at which ^i(x) diiidpziy) are not zero, we may write the equation thus:

q^iy) , P\{x) ~ dy ^ dx. p2{y) ^ ? i W

10 Some Types of Solvable Equations

The variables are now separated, since only^ appears on the left and only X on the right. Each member of the equation is now a differential, namely, the differentizd of its own quadrature. Since functions with equal differentials can differ by any constant, but cannot differ by more than a constant, we conclude that

J p2(y) J qi(x)

This is the general integral.

Example 1. T o integrate the differential equation

2{y - \ ] dx-\- x^ sinydy = 0.

This equation has^ = 1 as an integral. W i t h separated variables it is

- rfy = - - rfx. -y i x^

/* sin y 2 Thus / ^ dy = -+ y - \

is the general integral.


xydx - (* - 2) rfy = 0.

W i t h separated variables this equation is

dy x dx

x~2

Thus log> = 2 log jjv - 2) + X + A

is the general solution, with k as the arbitrary constant. By replacing each member of the last equation by its exponential, the relation is given the form

The differential equation also has the integral y = 0. Clearly all these integrals are included in the relation

, = 2 1 V ,

with the constant c arbitrary.

Exact Differential Equations 11

PROBLEMS

Find the general solution of each of the following differential equations.

i. {y -2\^dx - x^dy ^0. 2. y dx-\-dy ^ 0.

3. {x^+ \]ydy-\-x{y^ - A\dy ^0. 4. {2xV - r f x + 2xd> - 0.

5. sec* xsecydx sin ydy = 0. 6. y/1 ~ y^ dx -\-3 -s/\ - x^ dy - 0

7. + i 1 + x^ } (/y = 0. 8. |x - 11 cos )- rfy -= 2x sin > dx.

9. X log X flfy + -N/I + '^^ rfx = 0.

10. {1 - 2x y/l - sin^ x \ dx + V 2 - sin^ x dy 0.

11. xe"*dx -\-e~'dy = 0. 12. ^cos^^rfx -\-&ccydy = 0.

13. y/\ ]- y^ dx + xy dy = Q. 14. xy log y dx = sec x dy.

15. xf.?" + 4} dx = 0.

16. x\y -2\dx - \x - l]\y -\-A\ dy = {2 ~ y\dx.

17. x} dy = xy{a> - (fx!.

18. tan ydx-\- cot x rfy =0.

2.2. Exact differential equations The diflferential of a function F{x, y) is known from the calculus to be

Fx{x,y) dx + Fy{x,y) dy,

where and are the partial derivatives of F with respect to x and ^ . This is of the form of the left-hand member of equation 2.1. In at least some cases, therefore, a differential equation 2.1 is actually of the form dF{x, y) = 0. When this is so, the differential equation is said to be exict. Because a differential is zero only when the function is constant, the exact equation implies that

(2.2) Fix,y) = c.

This is its general solution. A differential equation 2.1 need not be exact. Before formula 2.2 can

be used, therefore, we must determine whether the equation is exact or not, and, if it is, how the function F(x, y) is to be found. We shall show how that may be done provided the functions Py{x, y) and Qx{x, y) are continuous.

If the differential equation is of the form dF = 0, it is so because

(2.3) P{x,y) = F,{x,y), Q{x,y) - Fy{x,y).


By partial differentiations, then,

Pyix.y) = Fx,y{x,y), Qxixyy) = Fj,,j:ix,y),

and, since the cross partial derivatives Fx,y and Fy.^ are equal, we see that

(2.4) Py{x,y) = Qx(x,>).

This is therefore a necessary relation for exactness. If it is not fulfilled the differential equation is inexact.

Suppose now, conversely, that relation 2.4 is fulfilled. Then let F{Xfy) be defined by the formula

(2.5) F{x,y) = r P{x,y)dx-{-R{y), J

in which R{y) is an undetermined function, XQ is any suitable constant, and the quadrature is made as though^ were a constant. The first equation (2.3) is then fulfilled, and also

Fyix^y) = \'Py{x,y)dx-\'R'{y).

But because of (2.4), we may write this last equation

FA>^.y) = r Q.{x,y)dx^-R'{y), J

namely, F^ix.y) = Q{x,y) - Q(xo,>) + R'{y).

The second equation 2.3 is thus also fulfilled \i R{y) is chosen so as to make R'ij) = Q(xo,>),that is, if

m = !Qixo,y)dy.

W i t h this evaluation, therefore, the function 2.5 fulfills both equations 2.3, and the general integral 2.2 is given by the relation

(2.6) /J P{x,y) dx-\- f Qixo^y) dy = c.

The method could have been applied with the roles of P and Q, and of X andy interchanged. W i t h a suitable constant the general integral is so found to be

(2.7) Q{x,y) dy + j P{x,yo) dx = c.

The two results 2.6 and 2.7 are equivalent. They may, however, be quite different in form, and the quadratures called for by one may be simpler than those for the other. It is therefore often wise to consider both formulas, and to carry through with the one that is found to be simpler.

Exact Differential Equations 13

Example 3. T o find the general integral of the diflferential equation

W -^y\dx-\-{x-\-^\dy = 0.

For this equation Py{x,y) = 1, and Qxix^y) 1. Condition 2.4 being fulfilled, the equation is exact. W i t h XQ 0, formula 2.6 gives the general solution

j^W^y]dx+ f e^dy = c,

that is, ^xy-\-i^ = c.

With^o = 0) formula 2.7 gives the solution

xy-\-e" - \ -hW =^c.

This differs from the one first obtained only by having {f: + 11 in place of c. That, however, is immaterial since c can have any value.

Example 4. T o find the general integral of the differential equation

6xy + 1

V v * - x^ dx + 3x^ -

2x dy = 0.

This equation is exact. W i t h XQ = 0, and^o = formulas 2.6 and 2.7 are, respectively.

0 6xy +

1

Vy' - x' dx = c. and

1

3x' -2x

Vv' - x-dy + 6x +

1

V l -x^ dx c.

T h e former is clearly the simpler one. From it we obtain the general solution 'hx-y + sin"^ \x/y'\ = c.

PROBLEMS Test the following differential equations to determine which ones are exact, and find

the general solution for each one that is exact.

19. {Ixy + 6x! dx [x"^ - \ \ dy = 0.

20. y 1 x"- y

1 X rfx + \x s dy = 0,

21. xy

3 2 ** xY - -1, ~ X 1

dy = 0,


22, 1

dx + 1

xy - ! dy = 0,

23, log > + -X

dx-V i - + 2> dy = 0,

24. \xy + log x} rfjf = (xy + log>l dy.

25. ^ sin y i/x + cos y dy = xe"^ dy dx.

x dx -\- y dy -26. ~ + {Un y -\-x] dx -\-X scc^ y dy = 0. x'

27. 1

( ' - : v ) ' 2 V ^ dy = 0,

28. {cos X tan ^ sin x sec ^ | dx + {sin x sec'' y + cos x tan^ y esc y] dy = 0 .

29. {x + > cos ( x y ) ^ " ' " " U x + {> + X cos (ry)*-"'f"'>}

Finding Integrating Factors 15

This , however, means that if equation 2.1 is multiplied by 0{xyy) it assumes the form dF = 0, that is, it becomes exact. Thus 0(x, y) is an integrating factor.

Suppose equation 2.1 is multiplied by ^ {F)d, where 4> is any integrable function. The equation becomes dF = 0, and, since the left-hand member of this is also a differential, the equation is again exact. Thus ^{F)d is also an integrating factor. Since the number of functions i> is unlimited, an equation 2.1 is thus seen to have an unlimited number of integrating factors.

Example 6. The differential equation

y dx X dy = 0

is inexact. Its general integral isy/x = c. Thus jy) is in this case y/x^y and, since P{xyy) = we see that 0{xyy) = \/x^. Thus 1/x^, and in fact ^{y/x){\/x~)y with any integrable function 4> is an integrating factor. W i t h appropriate choices of we see, for instance, that all the functions

xy y*- x' y^ X x^ y

are integrating factors.

2.4. Finding integrating factors The discussion of Section 2.3 has theoretical interest but little practical

value. The integrating factors it displays are expressed in terms of the general integral of the differential equation. T o find the integrating factors the general integral would have to be known. Then, however, there would be no need to concern oneself any longer with integrating factors.

There is no systematic, generally applicable, method for finding an integrating factor. If there were, all equations 2.1 could be solved, and our discussion of this equation could be terminated. The finding of an integrating factor therefore remains a problem. Sometimes one is detectable by inspection of the equation directed toward recognizing differentials in it. However, success in this detection is the exception, not the rule.

In the search for an integrating factor it is useful to observe that (i) A n y function of just one variable, multiplied by the differential of

that variable, is a differential; and (ii) A sum of differentials is a differential. W i t h these facts in mind we may try to collect the terms of a differential


equation into groups, each of which is recognizably transformable into a differential by some suitable multiplier.


{xy-\-2 log *} dx + {x^ + xe^\ dy = 0.

O n grouping the terms in the manner

x{x dy -{-ydx] + 2 log x dxxe" dy - 0, I

it is recognizable that the multiplier X/x transforms the first group and also the last term into a differential. A t the same time it leaves the remaining term a differential. Thus X/x is an integrating factor. It reduces the differential equation to d\xy\ + d {log;cj^ + rf^* = 0. The general integral is thus

xy + ( l ogx l^ + v = c.

M a n y schemes, such as the following, can be invented to help in the search for an integrating factor. Suppose there is an integrating factor of the form

(2.8) e{x,y) = dx+iHv) dv

The differential equation, after multiplication by (2.8), is then exact, and, by equation 2.4,

- 1P(X, V)J''

The Linear Equation 17

For this case, equation 2.9 is

- 1 + X + = a(x){2x + V ) -b{y)y{\ +x-\-y'^\.

The term 1 on the left must be matched on the right, and this suggests the trial of b{y)y 1, that is, b{y) = \/y. T o match the term B^ "^ on the left, we may then try a{x) = 1. These trials actually do fulfill the equa-tion. Hence, by equation 2.8, the function e^^^* which can be written, ye^, is an integrating factor.

PROBLEMS For each of the following differential equations, find an integrating factor, and, by

using it, integrate the equation.

31. ydx - {2x +y\ dy = 0.

32. {2xy2 -y^\dx-\r W - dy = 0.

33. \x^ - xy^'dy = 0.

34. (4*2 ~ xy -h 7x -\- 2y - I] dx + {x - 6y + \ ] dy 0.

35. [2x^y +>M dx + {2x^ - xy] dy = 0.

36. X sin y

1 ^ ) dx + X l o g X x^ COS y

dy = 0.

37. - sec y tan y X

dx {x sec y log x] dy ~ 0.

38. + dx + ~e " + JT sin ^ X y

dy = 0

2.5. The linear differential equation of the first order A diff"erential equation which is important both theoretically and prac-

tically is that of the form

(2.10) y' +p(x)y = q{x).

It is called the linear differential equation of the first crder because it is linear, that is, of the first degree, in y and y'. For an equation of this type an integrating factor is always easily found.

Let the equation be multiplied by ${x) dx. Its resulting form is

e{x) dy + {p{x)e{x)y - q{x)e{x)] dx = 0.

By equation 2.4, the last equation is exact if 6' = pix)6y that is, if 0'/6 = p{x). This integrates to give log $ = jp{x) dx, which is ^ = ^Jpix) di Thus tf/p^** dx is an integrating factor, by virtue of which the


differential equation becomes

The general integral is thus

^Jp(x) dx ^ J^(x)tf/P ^dx + C.

This method is easily remembered. It requires no memorization of details.


y' -\- y tan x = cos x.

After multiplication by d{x) dx^ the given equation is exact if 6' = 6 tan Xy that is, it 6 = sec x. Thus sec x dx is an integrating factor. The general integral found by its use is

y sec X =^ X -\- c.


{1 +x^]y'-{- {1 -x]^y = xe-'.

We begin by writing the given equation in form 2.10, that is, as

I -2x

1 +x' xe

y = 1 -\-x

The method then shows that 0{x) dx is an integrating factor if

e' = e 1 -2x

1 -\-x

The variables in the last equation can be separated to give

d0/e = 1 -2x

1 -\-x^ dx

Thus, log 0 = ;t - log 1 + x'^l, that is, = + x^). The inte-grating factor of the differential equation originally given is thus

e'dx/il 4 - x 2 ) .

By the use of this factor the differential equation is given the form

Changes of Variables 19

Its general integral is, accordingly.

- 1 1 2{1 + x M

PROBLEMS Find the general integral of each of the following differential equations,

39. >' + 2xy = e-^\ 40. y' + y cot X = 1.

41. / -xy

1 -x^ = 3x. 42. xy' log X + y - 2x2.

43. y' -X + 1

3x' + 3x'

45. y - y - ^ l + l o g x ! .

44, y' -\- xy = x.

46. y' + 1 + X

1 - 2x

47. y' 2x

= 1. 48. y' + 1 | - cot X X sm X

49. y' + y = 50. y' sin X + y sec x = cos x csc^ x.

51. / [ l + x M + y 2 - + 4x X X

52. Ix +4}y' + {x + 5)y = le'.

53. + 1 y _ 2xy = |x* + 2x2 + 1} eos X .

54. y' + 11 + 2x|y = {2x - \ \e-^'.

2.6. Changes of variables One of the best modes of dealing with a diferential equation to which

no other method seems applicable is to change one or both of its variables. By this means we may try to modify its form so as to make it subject to some known method of integration.

Let X and y be replaced by new variables s and , by formulas i

These formulas are to be reversible, so that s and can also be expressed i n terms of x and y. By virtue of these formulas

dx = fs {Sy u) ds + / u (j, w) du.

dy = g^, ) ds + gu(sy u) duy


and the functions P and Q are replaced by functions of s and , say,

P(x,>) = i ? ( j . a ) , Q(x,y) = r ( j ,) .

Differential equation 2.1 is thus transformed into

R[f, ds + / ^1 + T\g, ds + iu du\ = 0.

If this differential equation can be integrated, and has the general integral u, c) = 0, the general integral of the original differential equation is

obtainable from u,c) = 0 by replacing s and w by their values i n terms of x and)'.


+ 11 + > V ' l rfy = 0.

By the change of variables

X s, y e~'Uy

the given equation is transformed into

e-'u ds-\-\\-\- u'^We-' du - e-'u ds\ = 0,

that is, into u^ ds -\- (1 + u-j rfu = 0. This last equation is solvable since its variables are separable. Its solutions are u = 0 and

j - f (1/2^) - l ogu = c.

Since s = XfU ~ ^y, the original equation has the solution) = 0, and the general solution is

V2>-) - log > = c.


By the change of variables x = su, y u, the given equation is trans-formed into

-u^'du H- u^s^ + u]ds = 0.

We may drop the factor since u = 0 does not lead to an integral of the original differential equation. W e thus find that (du/ds) u = s^. For this last equation, which is linear, e~' is an integrating factor. The general integral is ue^" = { + 2j + 2) + Since s = x/y, u = y, the

Changes of Variables 21

general integral of the given equation is

X X

\y y

ASSIGNMENT 1. Show that the differential equation

(2.11)

in which d is a constant other than 1, is transformed into a linear difTercntial equation by the change of variable x " " Differential equation 2.11 is known as a Bemoulli equation, in honor of the Swiss mathematician James Bernoulli (1654-1705).

PROBLEMS Solve each of the following differential equations by use of the indicated change of

variables, or as a Bernoulli equation.

55. (1 + 2JC + 2y| + 12x + 2y - 1 - e] dy - 0,

56. 2 + 4* + 2> + 1

1 + x dx-\-{l + 2 x + > | < 6 ' - 0 .

X " u.

y ^ s 2u.

57. {2y - xylogx] dx -2x log xdy 0, X - J , y = u log s.

58. {3/ - x^y] dx + {x* sin y + x* - 3xy^] dy = 0,

59. \4xy - 2x* dx-\-dy0, |* ^ '^2

X = S/Uy y = s.

60. iy + V + x^y

1 dx+{x-\- x^y\dy = 0,

X = sm u, y = s esc u.

x '~ y 61. {sin~* y cos xj rfx H j==.dy ~ 0, AT = J ,

62. {y + V l - I rf^ + xrfy - 0, 1^ r V .

63. i2> + 2 x V -.-xiv

dx + |x +x'y} dy - 0, X = J ,

64. xV dx - {x - 2 / + 2y^] dy = 0, x " s

65. +


66. (^+'' + 1 - e^Hx-\-y-\- 1)J dx + {f*+>' - 1 + ^ " " ( x + ^ - 1)) dy = 0, x = {s + ) / 2 , y ^ i s - u)/2.

2 1 - 2 log X 67. / - > =

3JC 3xy

69. >' + ^ cot jt = CSC X .

3 >^

71 ' y jy + xy

75. / + -2)- = 8v'^ sin^ X.

68. + >

70. y + - ^ > = 2{x + l} v > . Jf + 1

"2- > + -> sin X = 2 2>

- . , , COS 3x

76. y' -\-3y\ogx = 3>^x Sin X cos X

77. / -Ay

X log X

I COS X

X log X

2x + 1

Ma

, 2x - 1 2\x^ - x + \\^ ^ 2{x^ - X + \\y

2.7. Equations witti homogeneous coefficients A function fix^y) is said to be homogeneous of the degree k if the effect of

replacing x and y by ax and ay, respectively, is merely to multiply the func-tion by a*. Thus

/(ax, ay) = a^f{x, y).

Example 14. The functions

t a n - " 1 .y/xy -yh / M o g 1 + -

X X y

are homogeneous of the respective degrees 0, 2 , and f . For

tan~" == a tan^"- ax X

{axp X

jayP^ log 1 + ax

ay = fl^-;;'^ log 1 +

A n y differential equation 2.1 whose coefficients P{x,y) and Qix^y) are homogeneous functions of the same degree is transformed by either one of

Equations with Homogeneous Coefficients 23

the changes of variable

(2.12) X = Sj y = sUf

or

(2.13) X = sUf y = u

into an equation with separable variables. The differential equation obtained by the one change of variable is ordinarily different from that obtained by the other. It is therefore often advisable to consider both changes, and to integrate the simpler of the transformed equations.


\y - Vx^ + / 1 dx- xdy = 0.

The coefficients of this equation are both homogeneous of the degree 1 T h e change of variable 2.12 transforms the equation into

V l -\-u^ds + sdu = 0.

This has separable variables, and has the solution

s{u + V l = c.

s = Xy and u y/xy the general solution in terms o( x andy is

y + Vx'+y' = c.


y'^dx-\- Ix^ - xy-\-y^] dy = 0.

The coefficients of this equation are homogeneous of the degree 2. The change of variables 2.12 transforms the equation into

[u + u^] ds+ {\ ~ u-^-u^sdu = 0,

whereas the change of variable 2.13 transforms it into

uds -\- {s^ -{- \ } du = 0.

O f these the second is obviously simpler. Its solution is

tan~^ J + log = c,

and, since s = x/y, and u = >, the general solution of the given equation is

tan"^ - + log;" = c.


PROBLEMS

Integrate the following differential equations.

79. {3x + 2y] dx + \2x +>) dy - 0.

80. {x < _ _ _ _ xy xY - 2x'y\dx + {2x* + xy* -\- xY + x*y] dy - 0

81. {x^y -{-xy^ - >'} dx + \xy^ ~ x^] dy - 0.

82. y sec^ * - dx + 2 y X sec -y

dy " 0.

83. \2x^y +/} dx + I V - 2je} rfy = 0.

84. \x -ye^f^ rfx + - rfy = 0.

85. - COS - tfAT X . y , y - s i n h COS -y X X

dy - 0,

86. M**^* ^ sin - | r f x + X ' - + x dy = 0,

89. fl>^^'rfx + {7 - flx^''*! dfy = 0.

90. >-s/x^ + / r f x - {x V * ^ + / +>M = 0,

2.8. Equations with linear coefficients

Every differential equation

(2.14) a-iX \- bxy ci\ dx + 02* + hiy + ^ 2 ! dy = 0,

whose coefficients are of the first degree in x and y, can be integrated. Such an equation may be of a type we have already considered. If h\ = 02 it is exact; if ci = C2 = Oj its coefficients are homogeneous of the degree 1; if 2 = 0, it is of the linear type 2.10. We may therefore suppose now that 6 2 ^ 0.

The change of variables

with an undetermined constant A, transforms the equation into

\A{s - k) -V Bu-VC\ds-\-udu = ^.

Equations with Linear Coefficients 25

If, here, = 0, the variables are separable. In that case we may assign k the value 0. \i A 7^ 0, we may choose h so that Ah-{-C 0. The equation is then one in which the coefficients are both homogeneous of the degree 1.


{2y-\]dx^-{ix-y + 2]dy = {i.

The change of variables, s = x -\- u = 3 ; f^ + 2, transforms the equation into

[ds-\-u ~ ()h-\-^\ ds ~ udu = 0.

W i t h A = ^, this is an equation with homogeneous coefficients, whose general integral is [u -\- 2s\'^[u 7>s\^ = c. In terms of the original variables, the general integral is thus \Sx y + 3j^j2jv I p = c.


{2x - 6> + 3) rfx - {x - ly ~ \ \ dy 0.

The change of variable, j = x + A, u = x 3^ 1, transforms the equation into {5w + 15 j ds -\- udu = 0. Here the variables are separa-ble. We may therefore take k ~ 0. The general solution, in terms of the original variables, is 2A: ^ log fx Sjy + 2j = c.

PROBLEMS

Find the general Intcgrsd of each of the following differential equations.

91. {2x+y - 2\ dx + [x - y-\-2] dy = 0.

92. {4x + 2> - l]dx -\- {2x - y\ dy = 0.

93. {7y - 3} rfx + {2x + 11 /y = 0.

94. {2x -y + 2,]dx - {4x - 2y -\- \] dy 0.

95. [2x + 3>! dx {ix - y - U] dy = 0.

96. I5x '\- 2y -\- \ ] dx {x - \ \ dy = 0.

97. {3x - 2> + 4} rfx - {2x + 7> - 1} rfy = 0.

98. {6x + 4> + 3i rfx + {3x + 2> + 2} rfy = 0.

99. { ; -+71rfx+ {2x+> + 3lrfy = 0 .

100. \4x-{-y - 2\ dx + {?>x+y - 2\dy = 0.

101. {5x -\-Ay ~ A] dx -\- {4x -\-5y - S\ dy ^ 0.

102. {2x-\-y - I] dx[2x - 9y - \] dy = 0.


2.9. Simultaneous equations

Just as in algebra a number of unknowns may be given by an equal number of simultaneous equations, so in the calculus a number of func-tions may be given by an equal number of differential equations. A pair of equations,

(2.15) z ' = / 2 ( x , : v , z ) ,

constitutes such a system for two functions > and z. The general solution of such a system consists of two equations in x, and and involves two arbitrary constants. T o obtain these equations two quadratures have to be made.

(i) A F I R S T I N T E G R A L . If the first equation 2.15 does not involve z, or if the second equation 2.15 does not involve y^ we may integrate the equation in question, and so obtain a first relation in x, y, and z. When neither equation 2.15 permits such a direct integration, we may use undetermined multipliers a , 3^, and 7 , to obtain from 2.15 the equation

(2.16) adx-^^dy^-'idz^ (a + i3/i + 7/2I dx.

If a , /3, and 7 can be chosen so as to make the left-hand member of equa-tion 2.16 a differential, say dQ (x, >>, z), and at the same time to mzJce the right-hand member a function (possibly zero) of 9 and x times dx^ this choice makes the equation appear as

(2.17) dB = F{e,x)dx.

The general integral of equation 2.17, (p{$, x, c\) = 0, yields one of the required equations

(2.18) *(x,>,z,^i) = 0.

(ii) T H E S E C O N D I N T E G R A L . T W O methods are available to obtain a second equation.

(iifl) It may be possible to reduce equation 2.16 to a form 2.17 by the use of new multipliers a, /3, 7 that are not proportional to those that were used in obtaining the first integral. A second integral of form 2.18 which involves an arbitrary constant c-z is then obtained.

(ii) It may be possible to eliminate either and^ ' , or z and 2', between equations 2.15 and 2.18. The liminant is then a differential equation for the remaining function. The integration of this liminant then yields the second integral.

Simultaneous Equations

Example 19. T o integrate the differential system

+ 2xz' + {2x + 2!e - 3 = 0,

xy' -y-{- x^z' + Ix^ -{-x}z - x = 0.

By algebraic means we may reduce this system to form 2.15

27

/ = - > - l , X X

The first of the last two equations does not contain z; hence it can be integrated directly to givey = x Cix^. By the elimination ofy between this equation and the second one of the reduced system, we find that

1 + 1

z.

the general integral of which is xz = 1 -\- ci{\ x] -\~ C2e general integral of the given differential system is, therefore.

X The

y X cix\ xz = 1 -\- ci{\ x] + C2e


y = xy + 2xz xy + 2xz

The choice of multipliers a = 0, |8 = 1 , 7 = 2 reduces equation 2.16 i n this case to dB = (d/x) dx^ with 6 = y -\- 2z. Thus a first integral 6 = c\Xy that is ,^ -\- 2z = cix, is obtained. The elimination oiy between this result and the second equation of the given system yields the equation

z' = 2z{cix - 2z\

ClX

T h e general integral of the last equation is ^ = cix^/{4x + C2), and the general integral of the given system is thus

y -\- 2z = ciXy CiX

z = 4x + C2


y - X z' = \

\y - x\z

The choice of multipliers a = 2>', 3^ = 2z, 7 = 2A: reduces equation 2.16 for this case to dB = 0, with B = y"^ z^ x^. Hence y'^ z"^ x^ = is a first integral. W i t h the different choice of multipliers a = x.

28 Some Typ)es of Solvable Equations

/3 = z , 7 = we obtain for (2.16) the form dB - 0 with B The generzd integral of the system is thus

^ xy - i z

~ - x^ = cu 2xy _ r 2 = C2.

PROBLEMS

Integrate the following differential systems.

103. xy' -\- z' -2z = Ixe', >' - r ' + 2z

104. \x^ + 4x\z' - U + 4}^ - y - y = 0, 3 x V - 3x* - xy' + ;r - 0.

105. y' - y - \x\\z' 2z {x + 3J, xy' - xy - ^ " X + 2.

106. y' y - xz' - z -\- x^ = 0, X

X X

1 X

X z - X** - x' - 1 = 0.

107. + -> - xr' - ^ + 0, { 1 + 2 log xjv' - xz' - z = 0. X

108. 2^ 5' + {3 tan X - 4|z = 3 sec X + sec* x, yy' - -fz' - 2z = -j sec* x.

109. / - 2 z 1

4yz + l I z

Ayz + 1

~4xz , 3x 1 1 0 . / = . z' =

111. y

3 + 4 y r

6x - ;y - 3 2y - 3z

3 + 4 ^

112. y = *'{2x -y] -yly

x\y + 2z\

z + 2 ~ Ax 2y ~3z

1 r = -2z)

x\y + 2zi

2.10. The linear differential systems with constant coefficient

The method of undetermined multipliers can be used to integrate any system

(2.19) >' = fly H- 6z + ^i(x), z'^hy-\-kz + g2(xh

in which a, bj k, and k arc constants. W e may in this case choose a and /3 and 7 as constants to fulfill the equations

= 0,

fljS + A7 = mfi, b^ -\- ky = my.

Replacement of a Differential Equation by a System 29

is possible i f m 5 taken to be a root of the quadratic equation

(2.20) a m h

b k m = 0.

T h e multipliers then give equation 2.17 the form

de= \me-\- ^ gi{x) + yg2(x)] dx,

with 6 = 72. Since this is a linear differential equation for 6, it has the integral

(2.21) {&y + yz]e-^ = J{^5iW + yg2{x)\e-^ dx + c^^.

process can be carried out for each value of m if equation 2.20 has distinct roots. Otherwise the elimination of ^ or z between (2.21) and an equation 2.19, yields a differential equation i n the other variable. T h e integration of this liminant gives the second integral.

PROBLEMS

Integrate the following differential systenss.

113. y' =ly - z' = 3y - 2z.

114. y = 2z - 4y, z' = -iz - 5y.

115. / = | y + 5r, z' = ~ h-

116. / - ~y + h, z' = ~h + h-

117. y = z, z' = -y -\- 2z.

118. y' = 3y - 4z, z' = 4y ~ Sz.

i\9. y' = 2y - z + e', z' = 4y ~ 3z-{-5.

120. y -'ly + Zz'h e^', z' = 2y - Zz - 7**.

121. y' -2y-\- 2^"'. = 3z + 6> + e~^.

122. y' = -2y -\-z-{' 3 " * ' , z' = 9y - 2z- 2e'.

2.11. The replacement of a single differential equation by a differential system

When a differential equation 2.1 is given, it is sometimes advantageous to regard x and y as functions of a third variable tj and to replace the differential equation by the system

(2.22) dx/dt = Q{x,y), dy/dt = -P(x,y).

The two equations that make up the general integral of this system express X and y in terms o( t. They therefore give a parametric representation of

30 . Some Types of Solvable Equations

the integral of the original differential equation. That integral in terms of X and y is obtainable by eliminating / between the pair of equations which constitutes the integral of the system.

Alternative to system 2.22 we could equally well replace differential equation 2.1 by the system

dx/di = k{x,y)Q{x,y), dy/di = -k{x, y)P{x, y),

with any chosen function k{x,y). The integration of this system is generally different for different choices oik{x, y). However, such different choices only determine effectively different parameters t. When t is eliminated the same integral for (2.1) is always obtained.


V l - / dx-\- {x + 2y] dy = 0.

Equivalent system 2.22 is in this case

dx/dt = x-\-2y, dyjdi = - V l - y-.

From the second of the last two equations we find a first integral relation

y cos \t + c\\.

W i t h y thus made known, the first equation of the system, which is linear in Jf, yields the second integral:

* = sin j ^ + (Ti i cos {( + (Ti I + c-i.

These two integrals together represent the solution of the given differential ' equation parametriccdly in terms of t. The liminant of t from them is

^^\-y'^-y-\-c COB y 1 with c = c^e This is the general integral of the differential equation originally given.

PROBLEMS

Integrate each of the following differential equations in terms of a parameter.

123. {2^ -y -2}dx+ - l*i dy = 0.

124. 13^ ^ + 2^ + 4} rf* + {1 + 2e-'\ dy = 0.

125. 1

2;- + -y

dx + (6* + 1 0 / - Ij rf> = 0.

126. 1 ! 7

dx + {* + 2y^ + 9} rfy = 0

Answers to Problems 31

127. 1

X dx - {2x - 2x^} dy = 0.

128. Xhx + 6x2} _ j _ _ 3y dx = 0.

129. 2 V ^ r f x + V 1 - * ((y -= 0.

130. rfjr - V l - ((y = 0.

ANSWERS TO ODD-NUMBERED PROBLEMS

1. > = 2 + 1 -hex

and y = 2. 3. >r = 4 +

5. sin^ y 2 tan x = c.

9. {y + \ / l ~ T ? " | logx

13. 2^ _ l o g x p - 1.

19. x^y ix^ - y = c.

23. X log ^ + log X + = c.

27. X + > X - >

+ sin ^ X = c.

1 X 1

^ r y

35. x-^y-^^; 4x^V^* - r 2 / 4

3x^^

1. y = log tan * x}.

11. fe-v'dy = c H - i ' i l - x j .

15. log ( . ' + 4 1 - { x - l - l ^ - ' + f.

17. > X + log {2 - x)2 + c, and> = - 1 .

21. Inexact.

25. e' sin > + xe~* c.

29. / ' ' < ' ' + i U 2 + / |

33. log X -X 3*

= c.

37. cos y; x sin y y log x

39. ye^* = X + c.

43. > = |x + I I U ^ +

47. y = 2x + f V^-

51. -^1x2 +x*| = x' + x + f.

55. X - > + Ix +^1

41. = c - {1 -x2}W.

45. :y - 0, and 2 = x^ - 1 +

61. X sin ^ ^ sin X + 63. log X - xV'* ' - f 65. 2 t a n - i Ix."} + y \/1 - >^ + sin"* > = c.


67. >* = log X + cx^. 69. y % 1

2 cos JK + c sin X

73. - ^ V T T x - V l -

75. 1

sin 2x + ^ tan x. 77. y - {/^" +rIogx}* .

79. 3x* + 4xy + - f. 81. I o g x + - + - - < : . y X

83. -5 + l o g \xy] - c. 85. ^ sin - c. X

x M x 1 87. -2 l o g - - -

J- I y ^ + log > 89. ae'^v + log > - c.

91. (2x +y - 2}* - 3 b - 2 ^ = c.

93. 7 log {2x + 11 + 2 log |7> - 3} = c.

95. [2x + 3>|2 - n{y+2\^ = c.

99. {> 4-7|2l3x + ^ + 11 - ff.

97. 3x^ - 4xv + 8x - ly"^ + 2y

101. 5x' +Sxy + 5y^ - 8x - \0y

103. y = fix* - 2C1X + 2 ^ + f2, = fi^:^-

105. r - < r 2 ^ - o U H - 2 | , z =r i {x*+x} - {x + 2 }

107. jf - ci log X , xz =- C i log X + ci (log x ) ' + C2 .

109. / - r + X " 1, y + z* - 2^.

111. X + 2; + 3z = c i , >z = X - X * + ^2.

113. y = cie* + ^2?"', -r = cie' + 3f2*~'.

115. > - c i ^ ^ * + C2e-'^^, z - - i V i ^ ' * - '^ 2'""'^ ^

117. y - cixe' + f2^, z = Cixe* + [ci + C2l*.

119. y =W - -k^ C2e-'^^ - W - + 5 + f i ^ + 4^2

121. > + z | i f - * - f^-" + 6;- - z = 6 ; - ' - x^"" + cze

123. ^ = 1 + C i * " * , ~ + f2*.

125. X =- 1 - cie-*^ + ^2^', = - i .

4x . - 8 *

1 127. - 1 + e i . - " , > + f 2 l ~ " .

129. X = i i 4 - [ci -

- ^ i d - / | V 4 - [ci - (1^ + 2 s in - 1

CHAPTER

3

Applications of Differential Equations of the First Order

3.1. Problems in velocities

T o solve a physical problem mathematically, it is commonly necessary to resort to a differential equation. The general integral of this differ-ential equation must be found, and from that the particular integral that fits some given condition must be determined. That is to say, the family of integral curves must first be found, and then the curve of this family that goes through a given point must be isolated. The differential equation expresses the relevant physical law, and the particular integral applies it to the specific situation.

For problems in velocities and rates, we shall use t to denote the time, and in the case of a moving particle we shall use s to denote the distance along its path. This distance is to be measured from some fixed point, with one of the directions taken as positive. The choice of the origin and of the positive direction may be made as seems most convenient. Once made, however, it must be adhered to throughout the course of the solution.

The velocity v and the acceleration a of a particle in a straight path are known from the calculus to be given by the formulas

(3.1) V = ds/dty a = dv/dt, a = v dv/ds.

The first two of these formulas are in effect the definitions of u and a. The third is obtainable as the quotient of the second by the first.

The fundamental law that describes the way particles move under the action of forces is Newton's second law of motion:

(3.2) / = ma, f = m dv/dt, f mv dv/dsy

these forms all being equivalent by (3.1). m stands for a constant called the mass of the particle, and / represents the component of the force acting in the direction of the path (or the sura of such components if there is more

33

34 Applications

than one force). Force / is to be taken as positive or negative depending on whether it acts to increase or decrease s.

When a body is allowed to fall freely without resistance under the pul l of gravity, its acceleration is the acceleration of gravity, which is denoted by g. This is the same for all bodies. Its numerical value depends upon the units in which distance and time are measured. In the foot-pound-second system it is nearly 32.2. Since law 3.2 applies to falling motion as well as to any other motion, we see that w = mg, where w is the weight of the particle. Thus

(3.3) m w/g.

W h i c h of the three forms of Newton's law 3.2 is to be used in any given case depends upon the variables in terms of which the problem is to be solved. If the problem calls for velocity in terms of time, the second form of (3.2) is appropriate, for the variables are v and t. If velocity is required in terms of distance, or vice versa, the third form is clearly the one to be used.

When a body falls in a resisting medium like air or water, the law by which the force of resistance is related to the velocity varies with a number of circumstances, such as the density of the medium, the shape of the body, the magnitude of the velocity. In each of the following examples we shall assume a simple law to take al l this into account. This would not neces-sarily be the law that experiment would bear out in the case of some other body. W e shall be solving differential equation problems, not establishing physical facts. For numerical results, we shall take the evaluation ^ = 32 to be sufficiently accurate for our purpose.

Example 1. A 2-lb. particle is dropped from a balloon at a great height. Dur ing its fall it is acted upon by air resistance. T o find the velocity of the particle when it has fallen 1000 ft., if it is assumed that the resistance amounts to kv^, with k = 1/20,000, in the foot-pound-second system.

The path of the particle is a vertical line. Along this we may take s as mccisured from the point at which the particle is dropped, and as increasing downward. The acting forces are then the weight w, which is positive, since it is directed downward like the increasing s, and the air resistance kv^, which is negative, since it opposes the motion and so is directed upward. The total force is thus {w kv^]. The problem calls for v in terms of s; hence we choose the third form of law 3.2, and obtain from it the differential equation

w kv^ = mv dvjds,

with w = 2,k^ 1/20,000, and, by (3.3), m = i V -

Problems in Velocities 35

The variables i n this differential equation are separable, integral is

,2

The general

- 6 2 5 log 2 -20,000

= s -\- c.

T o fit the condition ZJ = 0, when j = 0, the value of c must be c 625 log 2. The particular integral that applies is thus

- 625 log 1 -V 2

40,000 = s,

or, equivalently,

V = 200 V\ - e-'^^^^.

W h e n s = 1000, this equation gives v 178.6 approximately.

Example 2. A man with a parachute jumps from a great height. The ir combined weight is 192 lb . Dur ing the first 10 sec, before the parachute opens, the air resistance amounts to kv^ with A: = i n the foot-pound-second system. Thereafter, while 'the parachute is open, the resistance amounts to AID, with k = M. T o find the man's velocity of fall 15 sec. after the jump.

W i t h the direction of increasing s taken as downward, the acting forces are 192 and kv. The second form of law 3.2 thus gives the differential equation

192 - yti/ = 6 dv/dt.

which has the general integral

(3-4) log f l 92

I

V o

k-

fo

A n equation like 3.4 but with different values for k and 1 applies to each part of the fall.

For the first part of the fall A: = 4, and y = 0, when t = 0. The particu-lar integral that fits these conditions is

log 1 -V

256 8

that is, V = 256(1 -

A t the end of this part of the fall, that is, at / = 10, this gives the velocity

Pio = 256{1 = 182 (approximately).

36 Applications

For the second part of the fall k = 12, and, as we have just found, V = 182, when / = 10- The particular integral 3,4 that fits these condi-tions is

log V - 16

166 - 20 - 2(,

namely, = 16 + 166^^^"".

At t = 15, this last formula gives - 1 0 Pi5 = 16 + 166tf

The result is very nearly 16 f.p.s., since 166^~^^ amounts only to about 0.007,

PROBLEMS Solve each of the following problems, assuming that the all data given arc expressed

in the foot-pound-second system.

1. A ball weighing H lb- is falling. The air resistance amounts to u/lOO. If p = 15, at / = 0, what is the formula for v in terms of

2. A ball weighing 1 lb. is dropped from a great height. The air resistance amounts to p/32. Find the formula for s in terms of /, if f = 0, when j = 0-

3. A bullet weighing H o lb. is fired vertically downward from a balloon, with a muzzle velocity of 1000 ft./scc. The air resistance amounts to f^/1,000,000. Find the formula for v in terms of j .

4. A bullet of weight w is fired straight upward with a muzzle velocity of ro ft/.sec, The air resistance amounts to kv^. Find the formula for s (the height), in terms of c, while the bullet is rising.

5. In the case of the bullet of problem 4, find the formula for / in terms of v,

6. A certain particle weighs 8 lb. As it sinks in water its weight forces it downward, but an upward force, its buoyancy, also acts upon it. This buoyancy amounts to 4 lb. The water resistance amounts to v'. Find v in terms of j , if :J = 0, when j = 0,

7. When a certain body weighing 16 lb. is placed under water, it rises because of its buoyancy, which amounts to 20 lb. The water resistance amounts to Find the velocity with which the body rises / seconds after it is released.

8- A motor boat weighing 200 lb. is going at a speed of 20 ft./sec. at / = 0. The motor is then turned off. The water resistance amounts to -jj^ .^ Find the formula for D in terms of (.

9. A 1-lb. particle moves in a straight line on a smooth horizontal plane. A force acting upon it amounts to ^, and is directed so as to repel it from the point s = 0. Uv = Oy when j = 4, what is the formula for u in terms of s?

10, Find the formula for v in terms of s for the particle of problem 9, if the force amounts to l A ^ , and is directed so as to attract the particle toward the point j = 0.

Frictional Motion 37

11. A certain vehicle weighs 3200 lb. Its air resistance amounts to 2D. If a force of 40 lb. is applied to push it, and starts it into motion, find the formula for the distance s it has gone, when its velocity is v.

12. If the vehicle of problem 11 is moving with the velocity 20 ft./sec. at 0 , and a braking force of 40 lb. is thereafter applied to it, what is the formula for v in terms of

3.2. Frictional motion When a particle moves on a rough surface, its motion is opposed by

friction, which acts as a resisting force amounting to fif^fy where (i) fa is the sum of the components of the forces acting upon the par-

ticle in the direction perpendicular (i.e., normal) to the surface; and (ii) / i is a constant called the coefficient of friction. The value of M ts a

measure of the roughness of the surface. It is zero when the surface is perfectly smooth.

W i t h the friction taken into account, the motion of the particle is sub-ject to Newton's law 3.2.

Example 3. A 10-lb. particle is placed upon a plane whose slope is The coefficient of friction is and, in the foot-pound-second system, the air resistance amounts to v/4. The particle is at rest at / = 0, and is then released. T o find its velocity at any subsequent time /.

Let

38 Applications

that is,

PROBLEMS Solve the following problems, assuming all the data to be given in the foot-pound-

second system,

13. A boy and sled together weigh 80 lb. They coast on a hill whose slope is TTJ, and on which the coefficient of friction is -r^- The air resistance amounts to v/l'i. Find the formula for the velocity, i f = 0 , when ( 0,

14. What is the formula for the velocity of the boy and sled of problem 13 if the hill is perfectly smooth?

15- A boy on skis weighs 96 lb. He slides down a hill whose slope is which the coefficient of friction is -jfr- The air resistance amounts to 2v/l3. If p => 0, when J = 0, find the formula for s in terms of v.

16. A 16-Ib, particle slides up a plane whose slope is T, and upon which the coefficient of friction is x- The air resistance amounts to v/\0. If at / = 0 the velocity of the particle is 64, what is the formula for v at time t?

17. A 100-Ib- boy slides on the ice. The coefficient of friction is 7nr> and the air resistance amounts to v/20. If the boy's velocity at / = 0 is 40, what is it at time (?

18. A skater weighing 160 lb. allows himself to be blown along by the wind. The coefficient of friction is -j^ j-, and the wind pressure upon him amounts to 2 {30 v]. If his velocity is 14 when t = 0, what is it at any later time /?

19. In the case of the skater of problem 18, find the formula for s in terms of v.

20. A 4-lb. particle moves in a straight line on a horizontal plane whose coefficient of friction is T . The air resistance amounts to v^/l6. The force acting upon the particle amounts to 4J, and is directed to repel it from the point J = 0. If v = 0^ when J = 4, find the formula that relates v and j .

3.3. Problems in rates In many applied mathematical problems the immediately known facts

concern the rate at which a certain quantity changes. Finding the quantity itself then requires the integration of a diflferential equation.

Example 4. A certain mixing tank of 200-gal. capacity is filled with brine in which 60 lb. of salt are dissolved. Beginning at / = 0 the solu-tion in the tank is drawn off at the rate of 5 gal-/sec., and the tank is meanwhile refilled at the same rate with a solution that contains lb . of salt per gal. T o determine the amount x of salt that is in solution in the tank at time t.

Since at time / there is a total of x lb. of salt in solution, the amount of salt per gallon at that time is x/200. The rate of withdrawal is thus 5x/200. The rate of replacement is T ^ - We thus have the relation

Problems in Rates 39

dx 5A: 5 It ~ ~ 20 10*

This is a linear differential equation for x. Its general integral is x = 20 + ce^'^^^y and with c determined so that the condition x = 60 when / = 0 is fulfilled, the result is

X = 20-\- 40e-'^^^.

Example 5. A certain compound X is formed by the combination of 2 parts of a chemical U with 3 parts of a chemical W. When certain amounts of U and W are placed together, the rate at which X is produced is constantly proportional to the product of the amounts of U and W that are still present at the instant. T o determine the amount of X that is produced i n time t, if 10 lb. of U and 81b. of W are placed together at / = 0, and the amount of X at / = 1 is 2 lb.

Let x be the amount of compound X that is produced in time /. The amounts of U and W that have been used are ^x and %x, respectively. The remaining amounts are therefore 10 f x j , and 8 f x j . We have thus the differential equation

dx/dt = k{\0-%x]{B-%x].

Its general integral is

log 18 - | x l - log (10 - |x) = -^kt + .:.

W i t h c determined so that x = 0, when t = 0, this equation yields

log 200 - 15x 200 - 8x

^ - ^ k t

The fact that x = 2, when / = 1, serves to evaluate k. We find from it that k = l o g f I , and that therefore

.200 - 15x 200 - 8x

. 85 = / log ^ 92

This result can be written as

2oo{i - my\

15 - 8 (M)

PROBLEMS 21, A mixing tank contains 100 gal. of fresh water at / = 0. A solution containing

7 lb. of salt per gal. is then added to it at the rate of 3 gaL/sec, and the resulting mixture in the tank is drawn off at the same rate. Find the formula for the amount of salt in solution in the tank at time /.

40 Applications

22. In a certain volume of liquid 100 lb. of salt, but no more, will eventually dissolve. The rate at which it dissolves is always proportional to the amount that is still dis-> solvable. Find the amount x of salt that will dissolve in time /, if this amount is 5 lb. when 1 = 1.

23. In a certain volume of liquid, 200 lb. of salt, but no more, will eventually dis-solve. When a quantity of salt is placed in this liquid, the rate at which it dissolves is T^JTC times the product of the undissolved amount by the amount that is still dissolv* able. If 100 lb- of salt are placed in this liquid at / = 0, what is the amount x that is dissolved at time /?

24. From problem 23, what is the amount of salt that is dissolved in the liquid in time t if 200 lb. arc placed in it at ( = 0?

25. From example 5, if 10 lb. of chemical U and 15 lb. of chemical M^arc placed together at / = 0, how much of compound X is produced at time I?

26- From example 5, if chemicals U and were to combine in equal amounts to produce compound x, what would be the amount of compound produced in time interval / after u lb, of U and w lb. of IV were placed together?

27, Two chemicals U and IV combine in equal parts to form a compound -V. The rate at which the compound is produced is proportional to the square root of the prod-uct of the amounts of U and IV that are still present at the instant. If at ( = 0 there are 8 units of U and 2 units of W^ , find the formula for the amount of compound formed at time /,

28- If 2 parts of a chemical U combine with 3 parts of a chemical W and the rate at which the compound is produced is always proportional to the amount of U that is still present at the instant, what is the formula for the amount of compound produced in time ty if this amount is M at / =0?

29- The population ^ of a certain insect colony is at ( = 0, In the absence of outside influences the rate of its growth would be kp. The colony is, however, exposed to an influence that causes deaths at the constant rate . Find the formula for the population at any time t.

30. In the absence of outside influences a certain population p grows at the rate kp. However, it is exposed to an outside influence which induces growth at the additional rate {a +4^], where a and i are constants. If the population is at time /o, what is the formula for the population at time t?

31. The amount x of a certain chemical disintegrates at a rate which is always pro-portional to the amount that is still left. If it diminishes by 1 per cent in 10 years, and its amount is XQ at t = 0, what is the formula for x at time t?

32. The amount x of a certain chemical naturally diminishes, because of disintegra-tion, at a rale kx if left to itself. In use there is a further loss by wear at a constant rate . If the amount is XQ at time ( = 0, what is it at time t?

3A. Problems of flow A problem of flow is always based upon a differential equation. This

statement applies to a flow of water, or of air, or of some other fluid, and to an electric current, or the flow of heat from a cooling body.

Problems of Flow 41

When a l iquid whose volume is V is allowed to escape from its container through an opening at a distance y beneath its surface, the rate at which it flows through that opening is k V ^ , with some constant k. The value of this constant depends upon the physical circumstances of the individual Ccise, in particular upon the shape and size of the opening, the character of the fluid, the system of units used, etc. This rate of flow is also the rate at which the volume V diminishes. We draw from that fact the relation

(3.5) -dV/dt = k\fy.

This is not a differential equation such as we have already studied, because it involves too many variables. A differential equation in just two varia-bles is, however, obtainable from it, by substituting for dV its value in terms of y and dy. This value is determinable in any case from the shape of the volume F, and the location of the opening through which the l iquid escapes.

In a case in which the figure is simple, the formula for V in terms oiy may be easily discernible. A differentiation then gives dV. It is, however, often simpler to obtain the formula for dV directly from the figure without first finding V. This is done by applying the reasoning of the integral calculus, which says that dV is equal to the area of the surface of the l iquid multiplied by dy.

Example 6. A cylindrical tank of length 9 ft. and radius 5 ft. is mounted horizontally and is filled with oil . A t / = 0, a plug at the lowest point of the tank is removed, and a flow results to which relation 3.5 applies with k = T o find the depth ^ of oil in the tank at any time t while the tank is draining.

The circumference of the end of the tank is the circle + [y 5 j ^ = 25. For this circle, x = + The surface of the oil is thus a rectangle whose width is 2 V ' l O ) ' and whose length is 9. Since dV equals the area of this surface times dy, relation 3.5 yields the differ-ential equation

The general integral of this equation is

1 2 ( 1 0 - ; ' i ^ = ^ + r .

T o fit the condition that> = 10, when / = 0, the value o f t must be zero. T h e result can be given the form

42 Applications

t ^ V = 10 - ^ 180

In this case it was simpler to find the value of dV from the figure of the l iquid, than it would have been to find the formula for V and to differ-entiate it, for by elementary geometry V is found to have the relatively complicated formula

V = 9 25T - 25 cos"^ - 1^ + 0 - 5) VlO^ -

If the tank had been standing on end the situation would have been differ-ent, for then the formula for V would have been recognizable at a glance.

When a material body whose temperature is T is immersed in a quan-tity of water whose temperature is T', there is a flow of heat from the body to the water, or vice versa. The law that applies to this is Newton's law of cooling, which states that the rate at which the temperature of the body drops, namely, dT/dt, is proportional to the difference between the temperatures of the body and water. Thus

(3.6) -dT/dt = k{T ~ T'].

The constant k depends upon the physical values involved, among them the area over which the body is in contact with the water, the material of which the body is composed, and the units used.

T o obtain a differential equation in two variables from equation 3.6, the value of T' in terms of T must be substituted. This value can generally be found from the fact that the amount of heat which the body loses is the amount that the water gains. The change in the body's heat content is always obtainable as the product of its weight by its specific heat by the change in its temperature.

Example 7. A body weighs 45 lb. and is made of a metal whose specific heat is 5 . While at a temperature of 300 degrees, it is plunged, at ^ = 0, into 100 lb. of water at a temperature of 50 degrees. T o find the formula for the temperature T of the body during its cooling.

The amount of heat that the body has lost at time / is 45(g) {300 T]. The amount of heat the water has then gained is 100(1) j 501 (the specific heat of the water being 1). Since these two amounts are the same, we have

V-1300 - T] = 1001 r - 50},

whence 7"' = 65 {T/20). O n substituting this value into equation 3.6 we obtain the differential equation

Problems of Flow 43

-dTldi = k

T h e integral which applies is that for which T = 300, when / = 0. This integral is

r = -y*f|13 + 50 -^-t2i/20)fcj_

From this integral we can observe the temperature to which the body w i l l eventually cool. As t increases indefinitely the exponential term approaches zero. The l imiting value of T is therefore ^ ^ r ^ , that is, 61.9 degrees (approximately).

PROBLEMS 33, A cylindrical tank of length 10 ft. and radius 3 ft, stands on end, and is filled

with oil. When a plug halfway up its side is removed the oil Rows in accordance with formula 3,5, with k = ^ir- Find the formula for the depth of the oil in the tank while the tank is more than half full.

34, A tank is 6 ft. long, and its end is a square with a 4-ft. diagonal. At / = 0 it b half full of oil, and at that instant a plug in a lower edge is removed. The flow obeys law 3.5 with k = At what time is the tank drained, if it lies horizontally on one side?

35, At what time is the tank of problem 34 drained if it is mounted horizontally with a diagonal of its end in the vertical position?

36, A tank is in the form of a right circular cone with altitude 4 ft. and radius 5 ft. At ^ = 0, it is filled with water, and the water is then allowed to escape through a hole to which law 3.5 applies, with k = - j^ . Find the formula for the depth of water in the tank at time /, if the tank is mounted with its axis vertical and its apex downward, and if the hole is at the apex.

37- If the tank of problem 36 is mounted with its apex upward, and the whole is in the base, what is the formula for the time when the depthj^f water in the lank is^?

38. A tank is in the shape of a sphere with radius 4 ft. At / 0, it is filled with water, and an opening on the level of the center of the sphere then permits the water to escape. Find the formula for the time in terms of the depth k of the water in the tank, while the tank is more than half full.

39. A cylindrical tank with radius 4 ft, stands on end, and has an opening, for which k ^ "ffVj in its lower base. As water excapes through this opening, fresh water is run into the tank at the rate of 1 cubic ft,/sec. Find the formula for the time at which the depth of the water in the tank is A, if A = 9, when / = 0.

40. A 60-lb- piece of metal whose specific heat is -5- is at a temperature of 450 degrees. At f = 0, it is plunged into 300 lb. of water whose temperature is 60 degrees. Find the formula for the temperature of the body at time t.

41. A n 84b. body is made of metal whose specific heat is ^ T J . While at a tempera-ture of 308 degrees it is plunged into 11 lb, of water at a temperature of 53 degrees. Find the temperature to which the body eventually cools.

44 Applications

42. A 10-lb. body whose specific heat is -g- is at 0 degrees when t ^ 0. It is then plunged into 100 lb- of water at 80 degrees. If law 3.6 applies with k yf* what is the temperature of the body at time t?

43. A 5-lb. body whose specific heat is TnJ" is plunged at / 0 into 50 lb. of water. If the temperature of the body at this time is 200 degrees^ and if it eventually cools to 50 degrees, what is the formula for its temperature at time /?

44. A 50-lb. body whose specific heat is i^f, and whose temperature is 208 dc^ecs, is plunged into 20 lb. of oil whose specific heat is 3^ , and whose temperature is 50 degrees. Find the formula for the temperature of the body at time t?

45. The differential equation for the current i which flows under an e.m.f. (electro-motive force) in a simple electric circuit containing a constant inductance L and a constant resistance R is

(3-7) Li-Ri E. at

Find the formula for r\ if is a constant, and / = 10 when t ^ /Q.

46. From problem 45, find the formula for current 1 that flows in the circuity if I >- 0, when t ^ Oj and E ^ a sin

Locus Problems. Rectangular Coordinates 45

the point through which the curve is to go has a positive ordinate, we shall consider > positive too. If {x, y) is any point of the curve, the slopes of the tangent and normal lines are, respectively, y' and 1 /y ' , and these lines meet the x-axis at the points (x y/y\ 0) and {x + yy\ 0). The triangle therefore has a base of length [yy' + y/y' \. Since its altitude is we have

1 yy' +

This differenticil equation can be written i n the form y' V 2 y^ = y^ Its variables are separable, and the integral for which ^ = 3, when * = 1, is

- / -h 2 - V a i o g

This is the required equation. [V2 + ^]y

= x.

Example 9. T o find the equation of the curve that goes through the point (2,1), such that the area under any of its arcs and above the x-axis is equal to the length of the respective arc.

From the calculus we know that, for a curve above the x-axis, which extends to the right from a point (XQ, >O) the area under an arc is given by the integral / y dx. The length of the corresponding arc is

The curve sought is, therefore, one for which

By differentiation this becomes y + y'^j or, in another form, y' = + y/y^ 1. The general integral of this differential equation is

log \y + V / _ i j = x-^c,

and withe determined so that)" = l , w h e n x = 2, the result can be written

y + = ^-

It is possible to improve upon this form. O n transposing y^ squaring, canceling etc., we find that

We can now drop the ambiguous signs, since the two alternatives give the

I

I

46 Applications

same formula. The equation sought is thus

This is the equation of a catenary. It is the curve in which a rope or chain hangs when it is suspended from two of its points. In terms of the hyperbolic functions, which are defined in the calculus by the formulas

(3-9) svnhx - - r ^ ! , c o s h j c = i l ^ + ^""^h

the result can be expressed in the form^' == cosh (jt 2) .

PROBLEMS

49. Find the equation of the curve that goes through the point (1, 3), and for which the tangent at any point, and the Une joining that point with the origin, have slopes which arc the negatives of each other.

50, Find the equation of the curve that goes through the point (4, 3), and whose tangent and normal lines always form with the x-axis a triangle whose area is equal to the product of { -TT ) by the slope of the tangent line,

51, Find the equation of the curve that goes through the point (0, 1), and whose tangent and normal lines alw^ays form with the :r-axis a triangle whose area is equal to the negative of the slope of the normal line.

52. Find the equation of the curve that goes through the point (4, 2), such that the segment of its normal line between the curve itself and the >-axi3, is always bisected by the ^r-axis.

53- Find the equation of the curve through the point (1, 5), whose tangent and normal lines always make with the ^-axis a triangle whose area is equal to the slope of the tangent line-

54. Find the equation of the curve through the point (2, 5), whose tangent and normal lines always make with the >-axis a triangle whose area is equal to 4 times the slope of the normal line.

55. Find the equation of the curve through the point (-J, 0), whose tangent and normal lines cut from the >-axis a segment whose length equals the negative of the slope of the tangent line.

56. Find the equations of the curves through the point (3, 2), for which the tangent and normal lines always form with the and ^-axes a quadrilateral whose area is equal to 2xy.

57. Find the equation of the curve through the point (4, 2), for which the tangent line and the horizontal through the point of contact form with the x- and ^-axcs a trapezoid having the area 6.

58- Find the equations of the curves through the point (4, 3), for which the tangent line and the vertical through the point of contact always make with the x-axis a triangle having the area 9,

Polar Coordinates 47

59. Find the equation of the curve through the point (1, 1), for which each tangent line is at a distance from the origin that is equal to the abscissa of its point of contact.

60. Find the equations of the curves through the point (1, 1), for which any tangent line and the line from the origin to the point of contact make with the>-axis a triangle having the area ^.

61. Find the equations of the curves through the point (0, 5), for which the area under any arc and above the jr-axis is equal to 3 times the length of the respective arc.

62. Find the equation of the curve through the point (-ff, 0), along which the arc length and the function 2x^ increase at the same rate.

3.6. Polar coordinates Polar coordinates are sometimes better adapted to the statement and

solution of a given geometrical problem than are rectangular coordinates.

FIG. 3.

The two systems are related as shown i n F ig . 3. The connecting relations are

X = r cos df y r sin Q.

From the calculus, we know that the angle ^ and the differential of arc ds are given by the formulas

dB (3.10) t a n ^ = r I

dr and

(3.11) ds"" = r''dQ'^-\-dr''.

These are the essential formulas for dealing with the geometrical problems to be considered.

Example 10. T o find the equations of the curves through the point with the polar coordinates (1, ir/2) for which every normal line is a

Documents

Rudolph E. Langer a First Course in Ordinary Differential Equations