[M.V._Makarets,_V._Yu._Reshetnyak.]_Ordinary_diffe(BookFi.org)-2.pdf

Ordinary Differential Equations and Calculus of Variations

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ORDINARY DIFFERENTIAL EQUATIONS AND CALCULUS OF VARIATIONS

Book of Problems

M. V. Makarets Kiev T. Shevchenko University, Ukraine

V. Yu. Reshetnyak Institute of Surface Chemistry, Ukraine

World Scientific V h Singapore New Jersey London Hong Kong

Published by

World Scientific Publishing Co. Pie. Lid. PO Box 128, Farrer Road, Singapore 9128 USA office: Suite IB, 1060 Main Street, Rivet Edge, NJ 07661 UK office: 57 Shelton Street, Coven! Garden. London WC2H 9HE

ORDINARY DIFFERENTIAL EQUATIONS AND CALCULUS OF VARIATIONS Copyright 1995 by World Scientific Publishing Co. Pte. Ud. All rights reserved. This book, or parrs thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of materia] in (his volume, please pay a copying fee through the Copyright Clearance Center, Inc.. 37Congress Street. Salem. MA 01970. USA.

ISBN 981-G2-2191-6

This book is printed on acid-free paper

Printed in Singapore by Uto-Prinl

Contents PREFACE ix

1 F IRST O R D E R D I F F E R E N T I A L E Q U A T I O N S 1 1.1 Separable equations . . . . 1 1.2 Homogeneous equations 9

1.2.1 Quasi homogeneous Equations . . . . 16 1.3 Exact equations . 19

1.3.1 Integrating Factors . 2 5 1.1 Linear equations . . . 33

1.4.1 Bernoulli's Equation . 4 1 1.4.2 Darboux's Equation . . 44 1.4.3 Riccati's Equation . 46 1.4.4 Bool's Equation . . . . 50

1.5 Nonlinear equations . . . . . . . 52 1.5.1 Solvable Equations. General Solution . . 53 1.5.2 Solvable Equations. Singular Solution . . . 59 1.5.3 Unsolvable Equations . . . . . . . 6 2

1.6 Applications in physics . . . . 64 1.6.1 Mechanics . . . . . . . . . . 64 1.6.2 Hydrodynamics . 67 1.6.3 Electrical Networks . 68 1.6.4 Kinetic Theory . 69 1.6.5 Nuclear Physics . 7 2 1.6.6 Optics . . . . 7 2

1.7 Miscellaneous problems . . . . 74

2 N - t h O R D E R D I F F E R E N T I A L E Q U A T I O N S 77 2.1 Reduction of order . . . . . 77

2.1.1 Simple Cases . . 78 2.1.2 Homogeneous Equations . . 79 2.1.3 Exact Equations . 80 2.1.4 Linear Equations 82 2.1.5 The Initial Value Problem 83

2.2 Linear homogeneous equations . . . . . . 87

vi CONTENTS

2.2.1 Exponential Solution . . . 89 2.2.2 Power Solution . . . . 90 2.2.3 Transformations of Equation . . 92 2.2.4 The Initial Value Problem 94

2.3 Linear nonhomogeneous equations . 97 2.3.1 Method of Variation of Parameters . . 98 2.3.2 Method of Undetermined Coefficients 100 2.3.3 The Influence Function . 102 2.3.4 The Initial Value Problem . 103

2.4 Linear equation with constant coefficients . 107 2.4.1 The Homogeneous Equation with Constant Coefficients. 107 2.4.2 The Complete Equation with Constant Coefficients. Method of

Undetermined Coefficients. . . . 112 2.4.3 The Method of Variatiou of Parameters . . . . 120 2.4.4 Symbolic Methods . . 123 2.4.5 Laplace Transform . . . 131

2.5 Equations with polynomial coefficients 140 2.5.1 Changes of Variable . 141 2.5.2 Substitutions . . 1 4 3 2.5.3 Substitutions and Changes of Variable 145 2.5.4 Series Solutions 146

3 L I N E A R SECOND ORDER EQUATIONS 153 3.1 Series solutions . 153

3.1.1 Ordinary Point . . 153 3.1.2 Regular Singular Point 157 3.1.3 Irregular Singular Point 166

3.2 Linear boundary value problem . 172 3.2.1 Homogeneous Problem 173 3.2.2 Nonhomogeneous Problem 175 3.2.3 Green's Function . 178

3.3 Eigenvalues and eigenf unctions 182 3.3.1 Self-adjoint Problems 184 3.3.2 The Sturm-Llouville Problem 186 3.3.3 Nonhomogeneous Problem 188

4 SYSTEMS OF D I F F E R E N T I A L E Q U A T I O N S 191 4.1 Linear systems with constant coefficients 191

4.1.1 Homogeneous Systems . 191 4.1.2 Homogeneous Systems. Euler's Method 192 4.1.3 Euler's Method. Different Eigenvalues 192 4.1.4 Euler's Method. Repeated Eigenvalues 193 4.1.5 Repeated Eigenvalues. Method of Associated Vectors 194

CONTENTS vii

4.1.6 Repeated Eigenvalues. Method of Undetermined Coefficients . 198 4.1.7 Homogeneous Systems. Matrix Method . 199 4.1.8 Nonhomogeneous Systems 203 4.1.9 Method of Variation of Parameters . . . . 203 4.1.10 Method of Undetermined Coefficients . 204 4.1.11 Matrix Method . . . . 205 4.1.12 Initial Value Problem . . . 206 4.1.13 Laplace Transform . . . 207 4.1.14 Systems of Higher Order Equations 208

4.2 Linear systems . 216 4.2.1 Solution by Eliminations . 216 4.2.2 Matrix Method 219 4.2.3 Nonhomogeneous Linear Systems . . . . . 219 4.2.4 Initial Value Problem 221

4.3 Nonlinear systems . . . . . . 224 4.3.1 Method of Eliminations 225 4.3.2 Method of Integrable Combinations. 228 4.3.3 Systems of Bernoulli's Form 230 4.3.4 Method of Complex Variable . . . . . 231 4.3.5 Systems of Canonical Form . . . . . . 232

5 P A R 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 237 5.1 Linear partial equations . . . . 237 5.2 Pfaffian equation . . . . . 244

5.2.1 Mayer's Method. . . . . . . . . . . 2 4 6 5.3 Nonlinear partial equations . . . . 248

5.3.1 Lagrange - Charpit's Method . . 250

6 N O N L I N E A R E Q U A T I O N S A N D S T A B I L I T Y 255 6.1 Phase plane. Linear systems . . . . . 257 6.2 Almost linear systems . . . . . . . 266 6.3 Liapunov's second method . . 273

7 C A L C U L U S O F V A R I A T I O N S 279 7.1 Euler's equation 279 7.2 Conditional extremum 284

7.2.1 Isoperimetric Problem . . . . . . 2 8 8 7.3 Movable end points 292 7.4 Bolza problem . . 299 7.5 Euler-Poisson equation . . . . . . . . . 301 7.6 Ostrogradsky equation . . . . . . . 303

viii CONTENTS

8 ANSWERS T O PROBLEMS 307 8.1 Separable equations . 307 8.2 Homogeneous equations 308 8.3 Exact equations . . . - 310 8.4 Linear equations 312 8.5 Nonlinear equations 315 8.6 Applications in physics . . . 318 8.7 Miscellaneous problems . . . 321 8.8 Reduction o( order 323 8.9 Linear homogeneous equations . 326 8.10 Linear nonhomogeneous equations . . . . - 327 8.11 Linear equation with constant coefficients. . 330 8.12 Equations with polynomial coefficients 336 8.13 Series solutions . 338 8.14 Linear boundary value problems 342 8.15 Eigenvalues and eigenfunctions 344 8.16 Systems with constant coefficients . 346 8.17 Linear systems . . . . . . . 350 8.18 Nonlinear systems . . . . 351 8.19 Linear partial equations . . 354 8.20 Pfaffian equation . 355 8.21 Nonlinear partial equations. . . 355 8.22 Phase plane. Linear systems . . . 357 8.23 Almost linear systems . 358 8.24 Liapunov's second method . . . . . . 359 8.25 Euler's equation . . . . . . . . 359 8.26 Conditional extremum . . 360 8.27 Isoperimetric problem . . 361 8.28 Movable end points . . 361 8.29 Bolza problem . . . . . 362 8.30 Euler-Poissou equation . . . 362 8.31 Ostrogradsky equation. . . . . 363

B I B L I O G R A P H Y 365

I N D E X 369

Preface This problem book contains exercises for courses in differential equations and cal-

culus of variations at universities and technical institutes. It is designed for no n-mat hematics students and also for scientists and practicing

engineers who feel a need to refresh their knowledge of such an important area of higher mathematics as differential equations and calculus of variations. Each section of the text begins with a summary of basic facts. This is followed by detailed solutions of examples and problems.

The book contains more than 260 examples and about 1400 problems to be solved by the students, a considerable part of which have been composed by the authors themselves.

Numerous references are given at the end of the book. These furnish sources for detailed theoretical approaches, and expanded treatment of applications.

In preparing this book for publication, Mr. Y.-S. Kim rendered a great help to us.

ix

C h a p t e r 1

F I R S T O R D E R D I F F E R E N T I A L E Q U A T I O N S

1.1 Separable equations A differential equation which can be written in the form

M(x)dx + N(y)dy = 0, (1)

where M is a function of X alone and N is a function of y alone, is said to be separable. The solution is

j M{x)dx + j N(y)dy = C, (2)

where C is an arbitrary constant. The problem is then reduced to the problem of evaluating the two integrals in (2). In Eq.(l) we say that the variables are separated.

Example 1. Find the solution of the equation

y' = e'*>

which is such that y 0 when x = 0. The equation may be written as

y' = eV,

from which i t is seen that the separated form is

e~*dy e'dx.

Integrating now gives the general solution

-e~> = tz + C,

and we have to find the value of the constant C such that x and y vanish simultane-ously. On putting t = y 0, we have 1 = 1 + C whence C = 2.

The appropriate solution is given by

e"* = 2 - e'

1

2 CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS

Example 2. Solve the equation

xydx+(x+l)dy = 0. (3)

If y j 0 and x + 1 ^ 0, we can divide by tj and i + I and put the equation in the form

dy xdx

Integrating,

= 0.

J y J x + 1

M j r l + i - l n \x + l\ = C.

Taking exponential of both sides yields

j z - d ^ + l j e " 1 , C, = In |C|.

Equation (3) has also solutions y = 0 and x = 1 The first one can be obtained from the general solution when arbitrary constant C\ 0 and therefore JJ = 0 is the particular solution. The second solution x 1 can't be obtained from the genera! solution and therefore x = 1 is the singular solution. Then the solution of the problem (3) is

y = C,(x+ \)e-* if x jt - 1 ; also I = - 1 .

Example 3. Solve the initial value problem

y , c o t i + y = 2; y ( j ] = 0. (4)

KJf 2 and cot x ^ 0 the differential equation can be written as

dy

Integrating,

Whence

+ tan xdx 0.

f dy l sin xdx _ ^ I y 2 J cos x '

l n | s - 2 | - l n | e o s i | = C.

B = 2+&eosjf ) (5)

where C, = ]a\G\ is an arbitrary constant. To determine the particular solution satisfying the prescribed initial condition we substitute x = TT/3 and y = 0 into

1.1. SEPARABLE EQUATIONS 3

Eq,(5)> obtaining Ci 4. Hence the desired particular solution is given explicitly by

y - 2 - icosx.


y2 ( r 3 + l ) dx + (x3 - 5x2 + 6x) dy = 0.

If y j i 0 and x ^ 0, 2, 3 then the separated form is

dy (x3 + l)dx If

2 x3-Sx2 + 6x

or Sx2 - 6x + 1 x 3 - ^ + 6x } "

Using partial fractions we can write

5 i a - 6 i + 1 _ 5 i a - 6x + 1 _ A i 3 - 5 i 2 + 6 i ~ s.(g - 2)(x - 3} ~ x + x-2 ' 1 - 3 '

B C

Multiplying this by i we find

5 I 2 - fa + 1 ( i - 2)(z - 3)

Then letting i < 0, we have

5 i 2 - 6 i + 1

Cx = 4 + +-z - 2 g - 3

[x - 2)(x - 3)

Similarly, multiplying by I - 2 and letting x 2 yields

5x2 - 6x + 1 B =

* ( * - 3)

and multiplying by x - 3 and letting i 3 yields C = 28/3, Putting these values, we get

28 1 2 + 3 i - 3

dx j / 5 I 6 i 2 i -

which, on integration, gives

i l l 28 i = r + i l n i i | - ^ l Q | x - 2 | + - l n | i - 3 | + C, y 6 2 J

where C is an arbitrary constant. The given equation has also the singular solutions y = 0, (a; # 0, x * 2, a; # 3), s = 0 (y ^ 0), x = 2 (J j i 0) and x = 3 (y # 0).

4 CHAPTER I . FIRST ORDER DIFFERENTIAL EQUATIONS


(x 3 + \)dy - ydx = 0.

If i - 1 and g / O w t can write this equation in the form

dy dx

7 ~ * 3 + l '

Since i 3 + 1 = ( i + l ) ( i J i + 1), we have using partial fractions

1 A B x + C + i x + 1 + x a - x + r

where A, S, C are undetermined constants. Multiplying by (x + 1 )(x 2 - i + 1) we obtain

1 = A {x* - x + l ) + {Bx + C)(x + 1) = x'(-4 + B ) + x{-A + B + C) + [A + C).

Since this is an identity we have on equating coefficients of like powers of x,

A + B = 0t

-A-i-B + C = 0,

A + C = I .

Solving these we find A = 1/3, B - -1 /3 , C - 2/3. On substituting these values and integrating we have

f dy _ 1 / _rfx_ _ 1 / x - 2 / S 3 / x + l 3 / + l

= I | n | I + l l _ I f ~ ' / 2 )d (x - 1/2) 1 r dx 3 ' 1 3 / ( x - l / 2 ) J + 3/4 + 2 J ( x ~ 1/2) '+ 3/4

= ^ In |x + 11 - i In ( x ! - x + 1) + -L arctan + f j ,

or , I | 1 . (x + 1 ) 5 1 2x - 1 _ in ji = - In + - = arctan 7 = - + C,

6 x 1 - x + 1 ,/3

where C is an arbitrary constant. The original equation has also the singular solutions x = - l ( y ? ! 0 ) a n d V = 0 ( x ^ - l ) .

If a differential equation can be written in the form

y' = f[ax + 63,) (6]

1.1. SEPARABLE EQUATIONS

then we put z ax + by and have

or

in which the variables are separable. Example 6. Solve the equation

[x + y + l)dx + {2x + 2y-\)dy = 0.

If 2(x + j/) 1 ^ 0 the differential equation may be written as

, = ( + y) + i y 2{x + y) \

Put x + y z , so that

r = 1 + f ifx ( i i

Thus z is determined as a function of x by an equation of the form

( * z _ 1 _ z + l z - 2

which is separable

dx 2z - 1 2z-V

2 x - 1

or z-2

3

= dz, zjt 2,

02 dx. z-2

Integrating,

/K-H>*-/* 2z + 3 1 n | z - 2 | = x + C

from which, on putting z = x + j / , we have the general solution

2(x + j ,) + 31n|x + y - 2| = x + C,

or x + 2j, + 31n |x + 3 , - 2 | = C,

where C is an arbitrary constant. The original equation has also the singular solution z 2 or y = 2

CHAPTER I . FIRST ORDER DIFFERENTIAL EQUATIONS

Example 7. Solve

d By putting z = ix + 2y, we have

z' = 4 + 2y' = 4 + 2 V 7 ^ T

whence ,

4 + 2 V J ^ T Integrating,

/

1.1, SEPARABLE EQUATIONS 7

jsin^dy, + J + j e>dy3 = C,,

or y2 + y32 + ys = C

cosj/i + In \y2\ + e" = C2.

Sometimes system of equations can be reduced to the separable form on dividing by some function.

Example 9. Solve the system of the equations

y3y2 cosyidy, + yjy3dy2 + y2dy3 - 0,

yt_dyi + 2y3dy2 dy3 = 0.

On dividing the first equation by the y2y3 and the second equation by the y3/yi we have

cost^du, + y2dy2 + dy3 = 0, Vi 1

y'dys + 2yidy7 - -^dy3 - 0. 3

Integrating, we get dy3

Jyldyl + 2jy2dy2- j ^ = C2,

j cosyidy, + j y2dy2 + j y 3 - = Cu

' dy3

or sin 1/j + )ry\ + In \y3\ = Cu

~yi + yl + = CV 3 y3 P R O B L E M S Solve each of the equation in Problem 1 through 40.

2- 9 ~ yJuTxT) 3. y' y1 sin I . 4. xy' = v / n r F

5. y' = 1 + y2

6. xydy = y/y' + Idx. 7. (x2-l)y, + 2xy2 = Q; y(0) = \.


8. y' = 3y 2/ 3; j,(2) = 0. 9. *i' + y = y1; = 0.5. 10. 2x Jyy' + y 2 = 2. 11. y ' - x y 2 = 2iy.

(+$)* I - W < < ' - 14. e 'di - { 1 +e*)i,

1.2. HOMOGENEOUS EQUATIONS

(dy, + dy2)y2 = 0. 40. {y, + + dyi) + (y2 - y3){dy2 - fly3) + (y 3 - yt)(dy3 - dyt) = 0,

dy, + 5dy2 + 6dy3 + 7dy, = 0, Vldyt + j / ,dy 3 = 0.

1.2 Homogeneous equations An equation of the form

(!)

is said to be homogeneous whenever the function / does not depend on x and y separately, but only on their ratio y/x or x/y. Thus homogeneous equations are of the form

Function f(x,y) is called homogeneous of degree n if

that is, if i and y be replaced by ix and ty, i " factors out of the resulting function, and the remaining factor is the original function. For example,

* + 2 y \ s in*, h > ^ * . y x

x + y+Jxy, ^

are homogeneous of degrees 2, 0, 0, 1 and - 1 , respectively, since

(txy + 2(tyr = t> (x* + 2y*),

tx . X sin sin - .

ty y ] n ! 5 _ M y = ) n y

tx x tx + ty +


is called homogeneous if M(x, y) and W(at, y) are homogeneous functions of the same degree. Writing the equation (3) in the form

dy M(x,y) dx N{x,y)

the second member is now homogeneous of degree zero and so is a function of y/x.

i - ' (9-The form of a homogeneous equation suggests that it may be simplified by intro-

ducing a new variable, which we denote by u, to represent ratio of y to x. Thus

y = xv (4)

and Eq. (2) becomes

!-> Looking on v as a new dependent variable, we must consider u as a function of X, Differentiating Eq.(4) gives

dij _ dv dx dx

and hence Eq.(2) becomes x^ + u = F(v),

or d v

= . (5) F(v) v x ( S |

The variables x and v separated. Solving Eq.{5) and then replacing V by y/x gives the solution of the original equation.

Example 1. Solve the differential equation

' = - ( l a y - I i l + 1).

The expression on the right - hand side is clearly a function of y/x only.

The substitution y = xv gives

df v + *j7" = v()nv + 1),

or dv dx . = , v 0, v d 1.

ulnu x '

1.2. HOMOGENEOUS EQUATIONS 11

Integrating both sides yields

dti / d(lnu) r dx J u In u J In u J i '

la | In H | = l n | i | + In|C|.

This is equivalent to in |v | = Cx,

or v = e c :

Hence substituting for v in terms of y gives the general solution

y = xcc'.

v = 1 or y x is the particular solution corresponding C = 0. Example 2. Solve the differential equation

[* + 2i/)dx - xdy = 0.

The equation being homogeneous, we set

y = H I ,

dy vdx + xdv.

Then

{x + 2vx)dx - x{xdv + vdx) = 0,

if x ^ 0, the resulting equation is

(1 + 2v)dx - vdx - xdv = 0,

or (1 + v)dx -xdv = Q

whence , , dx dv x u + 1


l n | i | + ln |7| = ln

where C is an arbitrary constant. Hence we obtain

Cx = (0 + 1).

12 CHAPTER 1 FIRST ORDER DIFFERENTIAL EQUATIONS

Finally, substituting for u in terms of y gives the solution of the origin equation in the form

C l = (* + I I O I = 0

P R O B L E M S Show that the equations in Problems 1 through 30 are homogeneous, and find

their solutions. 1. {x-y)dx + (z + y)dy =


29. y- m - f - .

30. tfmm + pi y(0) = 0. Now consider the equation

dx \ o 2 x + i 2 y + Ci

which can be made homogeneous by change of variables. We put

x = I, + h,

y = yi + H then

a,! + b,y + c\ a\x, + b,yt + a,k + b,k + c 1 : a2x + i j y + cj = o j i i + b2yi + a2k + b%k + c2,

dt/i

14 CHAPTER I FIRST ORDER DIFFERENTIAL EQUATIONS


y - x - 4

Put x = Xi + k, y yi + k, and choose h and k so that

ft**-2 = 0,

-h + k-4 = 0.

These two equations give h 1, k = 3 and therefore X = tl 1, 1/ = i + 3. On substituting in the differential equation, we get

dyi _ x, + yi dx, j / i - X,

which is homogeneous and may be solve by substitution y, vx, Then we have

dv 1 + n v + x, -,

dx, v 1

or dv _ 1 - v2 + 2v

Xld7i~ v-i ' Separating the variable and integrating yield

Using partial fractions, we have

- I \ I 7~ 7=7 + " ?= \ dv = In I I I I + In ICI,

o r 1

- - (in \v- ( l + V2) | + In \v- ( l - y/2] |) = In [a, | + In \C\

which is equivalent to v2 -2v-l = Cx-2,

or

y\-2y,x, -x\ = C.

Since y, = 3/ + 3, i i = i + 1 the required general solution is y2 - 2xy - x2 + 4 i - &y = C,

v - 1 i/2 are the particular solutions corresponding to C = 0.



( 2 i - 4y + 6)dx + ( i + y - 3)dy = 0.

Equations (T) are 2h - Ak + 6 = 0, h + k - 3 = 0

whence h 1, fc = 2, and we put

a = i + 1, K = SA + 2.

We have

( 2 i , - 4 j f i ) d r 1 + (s, + ffOdj, = 0

which is homogeneous. Setting i/i = ux,, we have

{ 2 i i - 4ux,)dx, + ( r , + x,u)(u


35. (2i + y + l ) ( & - ( 4 : r + 2y-3)dy = 0. 36. { i - y - l ) + ( y - i + 2)K' = 037. ( i + 4y)y' = 2 i + 3y - 5. 38. (y + 2)dx = (2i + y - 4 ) d y .

, 37 - y + 1 4 1 " = 2 7 + 7 7 4 -

1.2.1 Quasihomogeneous Equations

A function f{x,y) is called quasihomogeneous of degree k if under the definite a and

/ ( C , ( * ) = * * / ( * , ) .

A differential equation

- / < * . ) W

is called fltiasinomoseneous if f{x.y) is quasihomogeneous of degree 0 a that is

/ ( * - * , t>) = i * - / ( * , f ) . (10)

For example, for

we have a = 2, 0 = 3, since

(t'z)*t*y x*y

The form of a quasihomogeneous equation suggests that it may be simplified by introducing a new variable, which we will denote by u

V

and equation (9) becomes separable. Example 5. Show that the equation is quasihomogeneous, and solve it

2xdy + ( * V + l ) ydx = 0. (11)


Writing this equation as

dy = y ( * V + l ) , 0 dx 2x '

we have

2 i Looking on / ( i , y ) as the quasihomogeneous function we must find a and /} in Eq.( 10). Then

tfWs^V+ 1) = _ a * V + l 2i"x 2z '

or

2 i 2x Whence 2ci + 4/J ci + /J = 0 o: or a+ 20 = 0. The equation being quasihomogeneous ( / a = -1 /2 ) , we set

and respectively

The resulting equation is

y = * - "

4 - / ^ dx 2 iix

dx 2 2x

du _ dx ~ 2*'

Separating the variables we have


dx _du = - 2 , u ^ 0 . x v"

k | x | + ln \C\ = -

Taking the exponential of both sides, we obtain

Cx = exp

Finally, substitution for u in terms of y gives the general solution of Eq.( l l ) in the form

Cx m exp

18 CHAPTER I FIRST ORDER DIFFERENTIAL EQUATIONS

Also x = 0 (y 7! 0 } and y = 0 ( 1 / 0 ) . A quasihomogeneous equation may be reduced to a homogeneous one by intro-

ducing a new variable ; y = f

where a to be chosen such that all the terms in the equation are of the same degree. Example 6. Solve the equation

2xy' (x - y 3 ) + y3 = 0-

Put y - .-" we have

J M T V ( I - z2a) + **" - 0. The first term 2x2oz"-'z' is of degree (or-1)+2 = a + l , t h e second term -2xaz3a-'z' is of degree (3a 1 + 1) = 3a and the last term z3a is of degree 3a. All the terms are of the same degree if a + 1 = 3a or or = 1/2,

On making the substitution y z l / 1 , we have

xz'i'h'{x-z) + z3'1 = 0.

This equation being homogeneous, we set 7 = ux and have

dz du u1 = 1 + U = d i d i u 1

whence du u

which on separation, takes the form

u - 1 . dx du = , u / 0.

u I


- t a | w | = l n | i | + ln |C| .

Putting y 3 / i for u we have the general solution of the given equation

y 2 = * l n | C V | .

Also y = 0. PROBLEMS Show that the next equations are quasihomogeneous, and find their solutions.

1.3. EXACT EQUATIONS 19

42. * V -43. 2 * V 4 xj,. 44. ydx + x(2xy + l)dy = 0. 45. 2y 4 * = 4Vy.

46. y' = y* - 1

47. 2*y' + y = y V * - * V

48. | i y j ' = x / i * - ! , " + y 1 . 49. 2y + {x2y + L).*y = 0 50. y ( I + iy)dx + ( l - zyjidy = 0. 51. ( i V + l ) y d i + ( 3 ; y - l ) i t f y = 0. 52. j>* - y^y'-xy = 0. 53. y ( l 4 V ^ j M ) + 2idy = 0.

1.3 Exact equations

A differential equation

M(x,y)dx + N(x,y)dy = 0 (1)

is said to be ernci if there exists a function, denote by U(x,y), such that

dU = ^-dx 4- ^-dy = Mdx + Ndy. ox oy

That is, if the first member of the equation is the exact differential of a function of x and y. The differential equation then takes the form

dU = 0,

and its solutions are defined implicitly by

V(x,y) = C,

where C is an arbitrary constant. Example 1. Solve the differential equation

-dx - ~dy = 0. v y

This equation is exact, since it may be written


and its solution is - = c. y

On the other hand the same equation in the form

ydx xdy 0

is not exact. T H E O R E M . Equation (I) is exact if and only if

BM _ dN dy dx (2)

The solution of the equation (1) is

J* M(x,y)dx + j " N{a,y)dy = C (3]

where a, b are any convenient constants.


2xydx+ (x 2 -y2) dy = 0 .

It is clear that dy 3x '

and the given equation is exact. If we take a 0, 6 = 0 several terms drop out and Eq.(3) gives

JB'2xydx + l" {-y*)dy = C, or

In solving exact equation (1) the student has various procedures available. The following example presents one of such methods.


e^dx -(2y + xe-')dy = 0.

Since

(-)--'=ih H % = - -the equation is exact. Then there exists a function f ( x , y) such that


Integrating the first of Eq.(4) with respect to x, holding y constant, gives

U(x, y)=j e'Ux + h(y) = xe~* + % ) , (5)

The function k is an arbitrary function of y, playing the role of the arbitrary constant. Now choose h(y) so that

-* From Eq.(5)

fr-x*-> + h>(y).

ou Setting - ~ equal to N and solving for h'(y) gives

ay

-xe-" + k'(y) = -2y-xe-".

Whence h'{y) = -2y.

Integrating % ) = V

The constant of integrating can be omitted since any solution of previous differential equation will suffice; we do not require the most general one. Substituting for h(y) in Eq.(5) gives

U(x,y) = xe-"-y2,

hence the solution of the original equation is given by

xe-" -y2 = C.


x (2x2 + y') + y (x* + 2y3) y* = 0.

H e r e dM , 3N

= 2 x y =

so the equation is exact. We have

U{x,y) = j x (2x2 + y 2 ) dx + / (y ) = | (s* + y V ) + f(y). dIJ

The partial derivative of the obtained function U(x, y) must equal y ( i 2 + 2t/2), dy

which yields yx* + f' = y(x2-r 2y2)


whence /' = 2y3

Hence

and the solution is

It is sometimes easier in finding the function U(x,y) to hold r constant and integrate N with respect to y.

Example 5. Solve

( x V * + 3 x V + " - l ) d i + ( x V + + y) dj, = 0.

This equation is exact since

| - ( * V + + 3 * V + - *) = ( x V + * + 3* V ) = ^ ( x V + " + y) .

Since the integration of (x'e*** + 3 i ' e ' + 1 ' ) with respect to i is more complicated, than ( i 3 e ' + " + JI) with respect to y, we use

hold i constant, integrating this with respect to y, gives

f - i V + + ^ + i ( t ) .

Here the function k(i) is to be found from

or ft'--*

whence

and the solution is


Exact systems A system of differentia] equations

is said to be exact system if

m . = ^ t i l i = 1 w t , = i

The solution of this system is

I*'

where y; g are any convenient constants. Example 6. Solve

( i + V2 + y^dyi + I ' + yi + ys)dyi-r ( i + i + &)%a = 0,

yzj/sdyi + yiVsdyi + y,y2dy3 = 0.

We have n 2

iW u = 1 + yj 4- y 3 , M w = 1 + yi + ya, Mis = 1 + y, + y 5 ,

M3, - yiy3, Mii = vm, ^23 = ifiyj.

Since

= 1,

= ys.

dMti 3MU dy2 dy.

SM, 3 dMi3 dy3 By2

dM2, dM22 dyi dy,

dM22 dM23 dy3 3y2

= yi-

the given system is found to be exact, thus putting yi 0 = o = ys o = 0 we get

F ds + ["(l+y,)ds + / " { l + y i + y ^

Jo Jo Jo

= 9i +(l + yi)yi + 0 + yi + y 2 )y 3 = C,,

and / y,y2ds = yiy ;ys = C2. Jo


Then the resulting solution is given by the following system

Vi + V + fa + Si3>i + ifij/3 + yjy 3 = C7i,

PROBLEMS Determine whether or not each of the equation in Problems 1 through 24 is exact.

If exact, find the solution. 1. (2 - 9 iy 2 ) xdx + (4y l - 6 i 3 ) ydy = 0. 2. ^dx + (y 3 + l n i ) d y = 0. 3. (2i + 3) + ( 2 y - 2 ) y ' = 0. 4. (2i + 4y) = ( 2 r - 2 y } y ' = 0. 5. ( 9 i 2 + y - l ) - ( 4 y - x ) f ' = 0. 6. (2xy2 + 2y) +


Determine whether or not each of the systems in Problems 25 through 27 is exact. If exact, find the solution.

25. (J/J + y 3)dy, + (y, + y 3)dy a + {yi + yi)dy 3 - 0, y-idy, + y,dy2 + y 3 dy 3 = 0.

26. * % + H-dto - ^ d y 3 = 0, 3/3 S3 yi

y2dy, + y\dy? + y|dy 3 = 0. 27. yiyt (2y2 + y3y


Example 7. Solve ydx xdy = 0.

We find that u, - and f i 5 = x/y3 are integrating factors. Then the general solution is

or l - c . X

In general case it is very difficult to find an integrating factor. We have for some simple cases:

a. If, in the equation

we have

a function of x alone, then

Mix + Ndy = Q

1 IdM dN\

u = exp

is an integrating factor. Example 8. Solve

Here

{xy- \)dx + dy = 0

N\dy d x ) ' 1

is a function of x alone. So we have the integrating factor

Then we have

exp ( y ) ( B " Udx + exp (~^dg = 0, (6)

BM f V \ dN lx*\

and Eq.(6) is exact. Integrating N = exp f^-J with respect to y, holding i constant

gives


The arbitrary function fc(jc) is so that

dU

or

and

3 i = y l e x p I ~ I ~ M = e x P I T I f 2 7 " -

= -exp ( y

= - J exp ^ y j d a .

Hence the solution of the original equation is given by

U(x,y) = i/exp ^yj - | exp (^J** = C -

Similarly , if 1 (8N dM\

is a function of y alone, then

H = n{y) = exp f(y)dyj

is an integrating factor.


(2xy2 - y) dx + (y 2 + x + y) dy = 0.

Here

M = 2xy* -y, N = y2 + x + y

1 (8N 3M\ 1 -4xy + 1 2 M \dx dy ) 2xy2 -y y'

Consequently 1

y2

Multiplying the original equation by 1/V gives

which is exact since

A (21 - - - ( l 3j/ ^ yj y2 dx \ y2 y)

28 CHAPTER J . FIRST ORDER DIFFERENTIAL EQUATIONS

And the genera] integral is (see Eq.(3)J

''\ + -\dy = C,

x2 - - + y + \ny = Ci y

where C, = C +t/o + lnj/o.

b. More generally, let p be a function of some function of x and y, w(x,y), let we say. Then if

dM dN dy dx ., ,

dt = dx dy

/(ui) is a function of to , then

ft - exp {^j f(io)dJj .

As an example, letting ul = x2 + y2 we have, if

dM 8N

2Nx -2My J \ x + y >

a function of the x2 + y2, then


\ xj y

Here M y , JV = . Assuming that fi is a function of w[,Jf) - I / J I e have

1 1

"1 - M - * - _ dy yi (-1)1 ' x\ ^


then

H = exp = exp In 1/

\ y / Multiplying the original equation by x/y gives

x --)dx + ^-dy = 0

and we have an exact equation

( . _ ! ) ( oy \ y) y a i \y

If we take a = 0 the second term in Eq.(3) drops out and integrating gives the solution of the original equation

Jo \ y


2y + 1

(x + y) dx+\3y + x +

x + J/)' dy = 0.

Here M = 2M + , ' v , , i V = 3y + x +

(* + y)2

Assuming that p is a function of x + y, we have

1 (dM dN

x + y)2

N-M \dy dx J x + y 1

then

ft = exp / - j d ( x + y) = e x p [ l n ( i + y}] = z + y. / * + y j

Multiplying the original equation by / i = i + y we get

1 2(*+) + z + y dx +

i x + y

dy = 0.

This equation is exact

dy 2(x + y)y +

x + y = 2z + 4y-

(x + y)2


1 d_ 3x

(3y + x)(x + y) + x + y

And the general integral is

r Jo

(3y + x)(x + y} + x + y

[i l dy + / -dx

Jxo X

or

= y3 + x*y + 2xy2 + In \x + y\ - In\x\ + In | i | - In | i D | = C,

;/3 + x2y + 2xy2 + In \x + y\ = Ci

where C, - C + \n\xa\. In simple equations it is often possible to group terms into combinations which

are exact and to solve by inspection. Example 12. Solve the differential equation

(x2 - sin 2 y) dx + x sin 2ydy = 0.

Multiplying by l / i 2 gives

xsm2ydy - sin 2ydx dx + = dx + d

sin2;/ = 0, x / 0.

Here each part is an exact differential and we have immediately

x + = C, also x = 0. x

Integrating factors are sometimes obtained by inspection. Certain combinations of differential suggests trial factors.


xy2dx + (x2y - x) dy = 0.

The equation may be written as

xy{ydx + xdy) - xdy = 0,

xyd(xy) - xdy 0,

from which multiplying by 1 jxy gives

1.3. EXACT EQUATIONS

The general solution is therefore

xy-\n\y\ = C.


(x2 + y2 -\-x)dx + ydy = Q.

The combination xdx + ydy = d([x1 + y2)

is suggestive, and we have

(x2 + y2)dx + 1-d(x2 + y2) = 0,

or multiplying by + j / 1 )


V v) * y PROBLEMS Show that the equations in Problems 28 through 37 are not exact, but become

exact when multiplied by the given integrating factor. Then solve the equations. 28. (y J + xy) dx - xHy = 0; f t = (*)"'. 29. (3xy + y1) + (x7 + xy)y' = 0; ft m at. 30. arV + x (1 + j , 1 ) y' = 0; a = ( i y 3 ) - ' .

/siny \ , / cosy + 2e - 1 cosr\ _

31. I - - 2 e - ' s m x \ d x + I 2 I = 0 ; i* = ye*

32. ydi + (2x-ye")dy = 0; u. = y.

33. ( l - dx + (2xy + ^ + *j dy = 0; u = i . 34. ( i 1 - i y 1 - y)dr + ( r 2 y - y 3 + i ) dy = 0; u, = ; u3 =

ry x' y' 35. (y - x)e-'dx + xe"*dy = 0; ft = e*.

36. (x - y)dx + f> + y)dy = I}; ** = u ( ^ 7 ^ )

37. (i - v)

1.4. LINEAR EQUATIONS 33

53. { z 2 - y 3 + y) dz + x(2y -i)dy = 0. 54. ( 2 i V + y)dz + {x3y - x)dy = 0. 55. y (z + v 2 ) dx + x 3 (y - 1 }dy = 0. 56. x(lny + 21nx- \ )dy - 2ydx. 57. (a 2 + l)(2xdx + cos ydy) - 2x sin ydz. 58. ( 2 z V + ( 2 z y - z ) d j , = 0. 59. I V + J + W - I ) ! / - 0. 60. ( i s m a + y cosci)dx + (y sin a z cos a)dy = 0. 61. (x ' -y )< i i + x(y + l)dy = 0. 62. y2[ydx - 2xdy) = x3{xdy - 2ydx).

63. (5 + l)rf*+^-l)*-0. 64. (x 3 + y ) d z - z d y = 0. 65. ( 2 z y 2 - y ) d x + (y 3 + x + y}dy = 0. 66. t V + > + ( V - ) * . M^y)-67. (y + z 2 ) dy + (x - xy]dx = 0, n = p(x2 4- y 3 ) . 68. ( r 3 + y ) d x - x d y 0. 69. (xy 3 + y ) d x - x d y = 0. 70. (2x3y*-y)dx + {2x1y3-x)dy = 0. 71. xy2dx + (x 2 y - x)dy = 0. 72. ( z 1 + y1 + 1)dx - 2xydy = 0,^ = fi{x2 y2).

1.4 Linear equations The general form of a first order iinear differential equation is

A(x)^ + Blx)y + C(x) = 0.

On division by the first coefficient, it can be put in the form

g + P(x)y = Q(x,. (1) If the second member is zero, Q(x) - 0, the equation can be solved by separating the variables

^ + />(x)y = 0, (2)

^ = -Pdx, ln |y | = - j P(x)dx + C,

y = Ctexp[-J P(x)dx\, (3) where Ci is an arbitrary constant.


ID general case (1} we multiply Eq.(l) by exp|/ P(x) ( x ) d x ] + e x p [ - / P(x)dx] j e x p [ | P(x)dx] Q(x)dX,

where C is an arbitrary constant. Example 1. Solve the differential equation

dy , 1 i dx x

Here P ( i } = 1/x, and we multiply by exp [/ P(x)dx\ = x,

^ , r

whence, integrating,

or y = -x3 + - .

4 x

Example 2. Find the solution of the initial value problem

j / - 2 x j , = i ; y(0) = l . Here P{x) = -2x,

j R(x)dx = - j2xdx M


Hence multiplying by e **, we have

e.-*(y'-2xy) = xe-*'

so that (ye-1)' = ."-

Therefore

ye"^= /are""1' rf* + & = - ^ e ^ + G

and finally

To satisfy the initial condition y(0) - I w e must choose C 3/2. Hence

1 , 3 J* y = - 2 + 2 e

is a solution of the given initial value problem. T H E O R E M i . Ify = U{x) is a particular solution oj (I) and ify = V(x) is a

particular solution of (2), then y = CV(x) + U[x) is the genera! solution of (I). We see from this Theorem that if we notice or can find in any way a particular

solution of (1), the problem is then reduced to the solution of the less complicated Eq.(2), the general solution of (1) can be written down at once.


y' + y tanx tanx.

We note that y = 1 is a solution. The equation

y' + y tan x = 0

can be solved by separating the variables

dy sin x ^ y cos x

or

In \y\ = In | cos x|.

Taking exponential of both sides yields

y cos x.

The general solution then is y = C cos x + 1.

36 CHAPTER I. FIRST ORDER DIFFERENTIAL EQUATIONS

T H E O R E M 2. / / y = U{x) and y = V(x) are different particular solutions of (1), then Ike general solution is

9 = G[V(*)-V(x)] + V(x).

This can be put in various forms. Since we have also

y Oj [U[x] V ( i ) j + U(x).


1 1 V T f = x - 1 x 1

This equation is found to be satisfied by y 1 and by y = x. The general solution is then

5, = (7( i - 1) + 1.

Consider the following method of solving the general linear differential equation of the first order (1).

If Q{z) is not identically zero, then we assume that the solution is of the form (3)

y = A(x)exp[- J P(x)dx\ (6)

where an arbitrary constant C\ is now a function of x and exp [ f P(x)dx\ is a solution of corresponding homogeneous equation. By substituting for t; in the given differential equation we have

A'{x)exp [- j P(X)dx\-A(x)P(x)exp [- / P{x)dx\

+P(xM(*)exp [- j P(x)dxj =Q[x), or

=


This equation can be solved by separating the variables

y x

Integrating,

/ = 2 / + l n | C | , J y ' x

or In |v| = 2in |s| + ln |C | ,

where C is an arbitrary constant. Hence taking the exponential of both sides, we obtain

y = C i ! .

Then we assume that the general solution of the original equation is of the form

y m A(x)x* (8)

Substituting (8) in Eq.(7)

x \lxA{x) + A'(x)x2] - 2A(x)x2 = 2x\

or A'(x) = Ix.

Integrating both sides yields A(x) = x2 + C1,

where C\ is an arbitrary constant. Finally, substituting for A(x) in (8) gives the general solution of Eq.(7) in the form

y=(x2 + C:)x2.

By a change of variable many equations can be reduced to linear form. We illus-trate this by some examples.


(2e* -x)y' = 1.

This equation is not linear. But it becomes linear if we consider x as a function of y. Then

dx _ _1_ dy~ dy_

dx and

+ x = 2e>. dy


The corresponding homogeneous equation is

+ * = 0, dy

dx , - -dy. x

Integrating In | i [ = - j , + ln |C| ,

or i - Cc-.

Then we assume that the general solution of the original equation is of the form

x = A{y)e->

By substituting for x we have

-AcT" + e " " ^ + Ae~* = 2e, dy

or

Integrating, A = e1" + ?,.

The general solution then is


2 ^ + j r W dx

This equation is not linear. Putting v y*, it becomes linear

dv _ dv

du x - + v = e 1 (9)

Solving corresponding homogeneous equation

dv i + u - 0 ,

dx

1.4 LINEAR EQUATIONS 39

dv dx V X '

= c X

and the general solution of (9] is of the form

A(x) v =

X

Then substituting in (9] we have

f * - 3 l * = - *

A' = e'. Integrating,

A = (*) = Q{x)e"

becomes linear after the change of variable v e~m*

= _ -m^L dx dx

and

m dx The proper change of variable in each case was suggested by the presence of a function of j ; and its derivative.

P R O B L E M S In each of Problems 1 through 50 find the general solution of the given differential

equation. Cxy' + 2jf = 3xty[G) = 0. 2. xy' + 3y = x1


3. y' + "y = 4. ( l + x , ) y ' - 2 x 9 = ( l + * 2 ) 2

5. if + 2xy = 2ze-''. 6. (2 i + l)y' = 4x + 2y. 7. y' + y tan i secx. 8. ( x y + c r)dx - xdy = 0. 9. r V + y + 1 = 0-10. y = x(y' - I C O S I ) .

11. 2x(x 2 + y)dx = dy. 12. ( x y ' - l ) l n x = 2y. 13. xy' + (x + l)y = 3 i 2 e-* 14. ( i + y 2)dy = yd*. 15. ( s i n^ + r c o t y l y ' ^ 1. 16. (2x + y)dy = ydx + 4 In ydy.

.7. r * - ^ -18. ( l - 2 x y ) y ' = y ( y - l ) . 19. y' + y = x 2 + 2. 20. y' - 3y = e3' + e" 3 1

21. = x<

22. y' = 2y + e*' + l . 23. xy' + 2y = (3x + 2)e3" 24. 2 i y ' - y = x 3 - x. 25. y' + y tanx = sin2x.

+ * i .

27. (x 2 - y 2 - l ) y ' = 2xy. 28. (y 2 - 6x) y' + 2y = 0. 29. ( x - 2 x y - y , ) y ' + y , = 0. 30. if - y = 2x - x 2

31. y ' - y = x - l ; y ( 0 ) = l . 32. y' + 3y = x 3 + l . 33. y' + 2y = x 3

34. y' ycotx= 2x x 2 cotx . 35. y' + y cosr = -sin2x. 36. y' + y = sin x + cos x. 37. x lnxy ' + y = 2lnx; y(e) = 0. 38. y' 2xy cosx 2xsin x; y(0) = 1. 39. y ' - 2xy = 1. 40. x lnxy ' - y = x(lnx - 1).

F x 2 + l x 2 + l


42. y' + ycaax = e " 1 .

43. y ' - 2 b S i n a * sin 2 i cos i

44. y' + xy = z 3; y(0) = - 2 .

45. y' + J U ^ .

46. y' + y tan i = xcos*i; 5/(0] = 1. 47. :r3dy - 2xydx = 3dz. 48. y'sin x - y = 2 sin J. 49. e**dy + feye? - i sin i ) dx = 0. 50. y ' i c o s i + yfxsin I + cosx) = 1. Find a particular solution by inspection; find a solution when the term not involv-

ing y is replaced by zero; and write down the general solution. 51. y' - y = 2. 52. y' + y = 2ex. 53. xy'-s = 1. 54. y' = y + l. 55. tf+y = * + !.

J.^.J Bernoullis Equation

The nonlinear differential equation

- J + P M ^ Q f x ) . , " (10)

where n is a constant but not necessary an integer, known as Bernoulli's equation, was studied in 1695 by the Swiss mathematician Jacob Bernoulli (1654-1705).

We rule out cases n = 0 and n 1, for which the equation is already linear. The substitution v = y 1 ' " reduces Bernoulli's equation to a linear equation. This method of solution was found by the German mathematician Gotfrid Wilhelm Leibniz (16461716)in 1696.

Equation (10] may be written in the form

Since

we get the linear equation

y-- + Py*- = Q(x).

1 dv _ dy 1 n dx dx

dV+P(x)v = Qfx). 1nix

42 CHAPTER J. FIRST ORDER DIFFERENTIAL EQUATIONS


J V + - f 3 = 0-

We have

jr di and, setting v y~2, we have

1 dv _jdj/ 2dx " dx

hence we obtain the linear equation

x1 du

Solving corresponding homogeneous equation

a 2 dw

we have dv _ ^dx v x '

or v = Cx*

Then the general solution of ( 10) is of the form

v = A{x)x*

Substituting for v in (11) gives

= 4 ^ + 4'**, dz

- Y [lAx3 + A'z*) + 2xAx" = 1,

or

Integrating,

where Cj is an arbitrary constant. Hence the solution of (11) is


Finally, substituting for v in terms of y gives the general solution of the original equation in the form

Practically to solve Bernoulli's equation it is helpful to use the substitution y = u(x)v(x) where v(x) is a particular solution of (10) if Q{x) = 0.


x V + 2*1/ - !/3 - 0.

Writing this equation in the form

X X

we have P(x) 2/x, Q(x) 1/x2 Putting y v.(x)v(x) we get

V U + U ( V + T ) = V " ( 1 2 ) Let v(x) is a particular solution of

2 t>' + - u = 0

x then separating the variables we find

= dx, v(x) = x " v x

On substituting v(x) in (12) we have

du dx u 3 a 6 '


* - * - * | * r * + $

OT I/O

Hence the genera! solution of the given equation is

, 7 \ -1/2


PROBLEMS Solve each of the following equations. 56. y' + Zy = a V . 57. ( i + l ] | y ' + y ! ) - y . 58. y' = y 4 cos x + y tan x. 59. xy'-2x1Jy = 4y. 60. xy' + 2y + x5yV = 0. 61. 2V'-X- = - p - .

y x' 1 62. y ' i 3 s iny = xy' - 2y. 63. ( 2 i 2 y l n y - i ) y ' = y. 64. y ' - 4 y - 2y^. 65. y ' - y + y 3 ( x 3 + x + l ) = 0 . 66. y' + y = xy1

67. y' + 2y = 2^/y. 68. l y ' + y = y 2 l n x . 69. y' + 2iy = 2zy3-70 3 y Y - Q y 3 = i + l . 71 y"~'{ay' + y) = x. 72. dx + (r + y 2)dy = 0. 73. (xy + x1y3)y'=\. 74. ydx + ( 2 x - 6 y 4 ) d y = 0. 75. y< = - J f - y .

x + y2

1.4.2 Darbouifs Equation

The equation

M(x,y)dx + N{x,y)dy + R(x, y)(xdy - ydx) - 0

where M[x,y) and N{x,y) are homogeneous functions of degree m and R(x,y) is homogeneous function of degree n becomes Bernoulli's equation after the change of variable y = xu(x). This equation is known as Durham's equation., after the French mathematician Gaston Darboux (1842-1917).


ydx + xdy + y2{xdy ydx) = 0.

Here M = y, N x are homogeneous functions of degree 1 and R = y 3 is homoge-neous function of degree 2. The equation being Darboux's equation, we set

y = 1 1 1 ( 1 ) , dy = xdu + udx.

Then xttdx + x(xdu + udx) + x2u2 [x(xdu + udx) xudx] 0,


or

2xudx + ( i 1 + x V ) du = 0.

Looking on x as the function of u we have

^x 1 " , , _ + s _ . _ r i i r j t ( , .

This is Bernoulli's equation. Putting x = z{u)v{u) where u(u) is a particular solution of equation

dv 1

or

We get

on substituting u = u ~ ' ' 2 we have

B - V V = _ | V *

or dz 1 - = - - J u .

Integrating both sides yields z-1 = u + G

whence

x - t r ' / ' f u + C ) - " 2

and on substitution u = y /x the solution of the original equation is given by

or Cxy + j 1 = 1, also i = 0.

P R O B L E M S Find the solution of the following Darboux's equations. 76. (x 2 + y 3 + y ) d x - x d y = 0. 77. (y 3 + 2xy 3)dy - 2y=dx + (x + y){xdy - ydx) = 0. 78. x 3 y 3 dx + x*y*dy + ydx - xdy = 0. 79. (x*y + y 3 ) dx - xydx + x 3dy = 0. 80. (x 3 - xy 2 ) dx + 2x*ydy - (xdy - ydx) - 0.


1.4.3 Riccatis Equation The equation

y'= P(x)y2 + Q(x)y + R(x) (13)

is called a Riccatis equation. Certain special cases of this equation were studied by the Italian mathematician Count Jacopo Francesco Riccati (1676 - 1754], In general, this equation cannot be solved by quadratures (that is, a finite number of integrations). If, however, some particular solution |ft('s) of this equation is known, we can obtain a more general solution containing one arbitrary constant through the substitution


y' = 1 + x2 - 2xy + y2

We find that y = x is a solution. Substituting

]

we have

dx v1 dx Simplifying, we have

Integrating, V = G - X,

where C is an arbitrary constant. Whence the general solution of the original equation is

PROBLEMS Find a particular solution by inspection; find the general solution of the following

Riccati's equations. 81. x2y' + xy + rV = 4. 82. 3 ' + y + 4 = G. 83. xy'-{2x + l)y + y2 = -x2. 84. y' - 2xy + y* = 5 - i 1

85. y' - 2ye* - y1 = e21 + e* 86. yr+y + y3=%


88. y' = 2 c o , ' - s i n ' s + y'_ 2 cos i

89. *+* = - J * 1 4

90. xy' = y 3 - 2xy 4 - 1 2 + 2 i - y. 91. x2y' = i V + xy + l. 92. x 3 y ' + (xy 2) 2 = 0. If we know two or three special solutions y y,(x) of

(13), the general solution is represented or follows. When y\(x) and y2(x) are the known solutions,

y-yi(x) = Cexp [J P(x)(y,(x)-yz(x))d: y - y*(x)

When jfi(z), yi(x), ys(x) are the known solutions,

y - y>(x) _ QPIX) - Vi(^)

Example 12. Solve y' 4- y 4- r* = 2.

We find that y, = 1 and y 2 = 2 are the particular solutions. Then the general solution is

^ = C e * p [ / , - 3 ) d t ] = C e -

or

y = -2 + ( d e " 3 1 + , also y = - 2 .

Riccati's differential equation

y' + ay1 = bxm

4k if m 0, 2, Trr where t is integer Is solved by quadrature. (In general, it is

(1 ~~ **) reduced to Bessel's differential equation by ay u'/u.)

a. If m = 0 we have y' + ay2 = b

or y = o - ay2

this equation can be solved by separating the variables

dy J . n f dy b ay2 J D


b. K m - 2 we have v+o*v=b

Putting y u/x we get the equation for ti

u' u

Whence

V + ax 2y i u ' u + a 5 = b.

tu' b + u au J

6 + u au !

The variables ii and x separated. 4 it

c. If m = , > 0 then the substitution 1 -2k

v - xm+3, y = 1

x'uju) ax (14)

gives du

O X 1 X 3 U X 2 U !

whence

vhere dv m + 3 m + 3 m + 3

-*+ ( m + 3)x+' m + 3 - bx"

m, -4 ( 4 - 1 }

(15)

l - 2 ( t - l ) By proceeding to do this substitution over and over again mt 0 can be obtained and we have the case m 0.

4k If m - , fc < 0 then the substitution

1 -2k

(16)

gives du b_ dv m + 1 -u m + 1


4(k + 1) where mi = - i ' , And the original equation is reduced to the case TJI = 0

1 Z\k 4* 1) again.


Here a = 1, 6 - I and m . From the equality ^ = ~~nT w e have t = 2 3 3 1 - 2fc

hence the substitution (14] we must make twice. The substitution (14) gives

1 1

and

y = - , pJl* I ' U x

* S2 TJ3U 2 ^ du 3 Then the original equation takes the form

^ + 3 a = - 3 o H (IT) du

4it Here a 3, 6 = 3 and n i | = 4 and from the equality 4 = we have now

On making the substitution

1 _ l ' Z V , u = - + U 3 U J 3u

we have , 2 1 ^ 1 du, 1

U 3 U J 3u 2 u 2 iu 2 dz u 2

and (17) takes the form

, i 1 dui 1 2 < + 3u 2 = - r y - r - r - -

1 . 2 1 . 1 \ 1 du> . 3 t u ^ U I 3 3 U 3 U I 9 u 2 / U 1 ! ! ) 2 dz U * U I 2 '

= - 3 ( l w ) .

Separating the variables and integratiog both sides yields

3z + C = arctati-u/,

so CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS

where C is an arbitrary constant. Finally, substitution for z and in in terms of y and x gives the solution of the

original equation

Tn C = arctan jz r: rz x-'/3 [x"3 - \x(xy + 1}

PROBLEMS Find the general solution of the Riccati's equations 93 V+V = 2*~2 94. 4y' + y 2 = -ix-1

95. S ' + y J = 2 i - <

96. y' + y 2 = x " 4 / s

97. y' + y2 = - 2 - " . 98. y' - y1 = 2x~2'3

1-4-4 Bool's Equation

The equation

1 1

is known as Bool's equation, after the English mathematician George Boole (1815 1864). The substitution v 1", y = uv where u = tt(u) reduces Bool's equation to the form

au' + W 2 = Ci," 1

where m = 2 and therefore in the case when n = , t = 0, 1 , - - is solved 1 2k by quadrature.

Example 14. Solve 1 . 1 a

It is Bool's equation, and we rewrite It in the form

" 2-* 3 * 2

Here a = -1 /2 , 6 = 1/2, n = 1. Putting 0 - i " 1 ' 2 , y = u r - " 1 , where u = u{v) we obtain

^ d 1 _t/t\ du 1 _ 3 / j _ l / 3 d u d t i 1 _ 3 / , = [ux ' I = 1 ' m ' x ' -; U J : ' dx dx \ ' dx 2 dv dx 2

2 ' 2


Substituting this into the given equation we get

or

it-*-- (" We obtain the Rieeati equation where a = 1 , b = I , m = 4 and it = 1 then substitution (14) gives

- i 1 1 Z V , u = ,

v'w V , _ 1 dm 2 l_ _J_d^dz__2_ _1_ V V 2 U J 2 du u3ui u a v 2 u i J dz du u 3iu v1

1 d u l 2 1 u 2 iu 2 dz U 1 U31H u 2

and (18) takes the form

1 dm 2 1 / 1 1 \ J

or

u - u = r r - , ,

ur* ( l + u.*).

Separating the variables and integrating both sides gives

z + C = arctan w,

or tan(C + z) - w.

By substitution for 1 and tu in terms of y and 1 we obtain the general solution of the original equation

= - l + * " * a t ( ( 7 + * " ) .

P R O B L E M S Find the solution of the following Bool's equations 99. xy' - 5y - y 2 = x1

100. 3xy' - 9y - y 3 = x1'3.

101. y ' + - ^ y + -^y3 = i -

3 y a

102. tt' + - y + y = i . By differentiating it is possible sometimes to reduce integral equation to differential

one. We illustrate this by the following example.


Example 15. Solve the integral equation

F[x - t)ydt = & + f''y(t)dt. Jo Jo

Differentiating both sides yields

2 + y(x) = f'y(t)dt. Jo

Putting x - 0 we find the initial condition y(0) = - 2 . Differentiating once again gives

y = !/

Separating the variables we have

y integrating now gives the general solution

- Ce*

and we have to find the value of the constant C such that y(0) = - 2 . On putting x = 0, y 2, we have - 2 = C. The appropriate solution is given by

y = -V

1.5 Nonlinear equations There is a first order differential nonlinear equation of the form

F(x,y,P) = 0, (1)

where p = y'{z) and F is a function of three variables. T H E O R E M There exists a unique solution y - y(x), x 0 - h < x < x 0 + k

(where h is sufficiently small) of Eq.(I) that satisfies the condition y(xg) yt, for which y'{x0) jfi, where yj is one of the real roots of the equation F{x0,y0,y,) = 0, if in a ctosed neighborhood of a point (xa.ya, Vi) the function F possesses the properties'

1) F(z,y,p) is continuous in all the arguments; 2) the derivative F^(x ,y,p) exists and is nonzero; 3} there exists the derivative \F^x,y,p)\ < M. Any solution of Eq.(l) of the form y = i^(x, C) in which C Is an arbitrary constant

is called a general solution of the equation. Any solution that may be obtained from the general solution of Eq.(l) by assigning particular value to the constant C is called a particular solution of that equation. A solution of Eq.(l) that cannot be obtained

IS. NONLINEAR EQUATIONS 53

by assigning specific value to the arbitrary constant in the general solution is called a singular solution of that equation.

In order to solve of Eq.(l) one should first investigate 1) whether i t is solvable for a variable (for example: F(x,y, p) = y-\n{xyl)+x2y' =

0, and then y Infij / ' ) - xiy'}\ 2) or it is unsolvable for every variable (for example F(x, y,p) xy \n(xrf) +

siniyyO = 0 ) . Then in the first ease one should investigate 1) whether all obtained functions are real; 2) or some obtained functions are complex. Finally, in the case 1.1 the following cases are possible: 1) Eq.(l) is solvable for j/f/x); 2) Eq.(l) is solvable for y{x); 3) Eq.(l) is solvable for x. In general, the cases 2) and 1.2) are beyond the scope of our text.

1.5.1 Solvable Equations. General Solution

I . E q . ( l ) is solvable for y'{x). Solving Eq.(l) for p leads to:

=*/*(%!/), = 1, 2 , . . . , i , (2)

where I is a number of solutions. Further we must solve each of the I equations. A general solution of Eq.(2) at any fixed ; can be found in any of the following forms, namely: an explicit function y y(x,C); an inverse function x x{y,C); a parametric function x = x(t,C), y = y(t,C), where t is a parameter; an implicit relation 0. Proceeding as stated above we obtain y' 2^/xy and y' 2^/xy Upon

integrating we find

m =

and j y(x)={~Yl2 + Ci) ,

where d are arbitrary constants. Putting Cj = - C ] s C we obtain the general solution of the given equation


The given equation also has the singular solution y = 0 [see below], a) At fi(x, y) = f,(y) Eq.(2) becomes separable and then

dy (2a)

where C, are arbitrary constants. Example 2. Find the general solution of the differential equation

y'2 - 4y = 0

at y > 0. According to Eq,(2) we find y' 2^/y, y' = -2^/y. Hence solutions of the given

equation are inverse functions x ^/y + C, and x y/y + Cj , where C; are arbitrary constants. Putting Ci C\ = C we obtain the general solution of the given equation

y = {x + C)2

Here y = 0 is the singular solution.

b) At fi(x,y) Eq.(2] becomes separable and then:

y = j ft{x)dx + C (2b)

where C, are arbitrary constants. Example 3. Find the general solution of the differential equation

y'2 - 4x = 0

at x > 0. According to Eq.(2) we obtain y' 2^/x, y' -2y/x. Hence solutions of the

given equation are

and

where C, are arbitrary constants. These solutions can be written as a general solution in the form

where C is an arbitrary constant. 2. E q . { l ) is solvable for y. Solving Eq,(l) for y leads to

y =

1.5; NONLINEAR EQUATIONS 55

where f is a number of solutions. Further we must solve each of the f equations. Let p{x) y'(x) is a new function of x alone. Differentiating Eq.(3) with respect to x then yields

p(x) = C(*.P) + v^(x,p)p'(x)-Hence

^ ^ S f 1 ' " ' This differential equation is completely analogous to Eq.(2). If an explicit function p = p{x, C) is a general solution of Eq.(4), then by substituting this function in Eq.(3) one find a general solution of Eq.( 1) as an explicit function y w(x,p(x, C}). If an inverse function x x(p,C) is a general solution of Eq.(4), then this function together with Eq.(3] give a general solution of Eq.(l) in the parametric form x x,(p,C), y ipi(xi(p,C),p), where p is a parameter. There is a similar case when Eq.(4) has a general solution in the parametric form x Xi(t,C), p = pi(t,C), where t is a parameter. If an implicit expression ipi(x,p,C) = 0 is a general solution of Eq.(4), then this expression with Eq.(3) give a general solution of Eq.(l) as an implicit parametric relation y y,(x,p), (I>j(x,p,C) 0.

Example 4. Find the general solution of the differential equation

According to Eq-s (3) and (4) we find

m - x (x 2 + 2p)'

By substituting p x2/z, where z is a new unknown function, we obtain, after some minor calculations,

z[2z + 3) z + 2

Using partial fraction and integrating gives

- l n | z | - - l n |2z + 3| = ln|Cx|, 3 6

where C is an arbitrary constant and 2z + 3 ^ 0, z / 0. Now we obtain, recalling the substitution for p,

3 C V 1 ~ 1 - 2 C V '

Hence the general solution of the given equation has the parametric form

_ 3 V _ x ~ 1 - 2 C V "


P 2 y = x + I p '

where p is a parameter. The singular solutions are y 0 and y 2 i 3 / 9 . a] At

1.5. NONLINEAR EQUATIONS 57

Example 6. Find the general solution of the Lagrange equation

V 1 - v' - y = 0.

According to Eq.(4b) we obtain

dx+2x 1 dp p - 1 p ( p - l ) '

if p ^ 0 and p f I . Solving it (see Sec.1.4) we find

*m = ( p - 1 )

where C is an arbitrary constant. Hence

_ C + p + In |p[

* ~ ( P - D J '

y = i p 2 - p

is the general solution of the given equation. If p = 0 or p = 1 we find two singular solutions y = 0 and y = 1, respectively.

c) Clairaut 's equation. An equation of the form

i-n'+fW (4c)

is known as a Clairaut's equation, after the French mathematician Alexis Clairaut (1713-1765). Notice that this is a particular case of Lagrange's equation, if 4i(y') y'. Hence the condition p 0(p) = 0 is always fulfilled. In this case Eq.(4) can be written in the form

(* + /< tP) ) f - f t

Hence the general solutions of the Clairaut equation is

p - C, y = Cx + / ( C ) ,

where C is an arbitrary constant. Moreover there exists a singular solution of the form

* = - / ' ( P ) . S/ = * P + / ( P ) ,

where p is a parameter. Notice that the singular solution is the envelope of the family of integral curves defined by the general solution.

Example 7. Find the general solution of the Clairaut equation

y xy' + e"

.58 CHAPTER ) . FIRST ORDER DIFFERENTIAL EQUATIONS

Since j{p) exp(p), we hare the general solution

y = Cx + cc,

where C is an arbitrary constant. The singular solution is

x - y - xp + ep,

where p is a parameter. 3. E q . ( l ] is solvable for x. Solving Eq.(l) for x leads to

x = Mv-y'). i - M t, (5)

where / is a number of solutions Further we must solve each of the / equations Let p(y) Jf'(x) is a new function of y alone. A differentiation of Eq.(5| with respect to i then yields, taking dp(y)/dx pp'{y) into account,

Then

This differential equation is completely analogous to Eq.(2). If an explicit function p{t/, C) is a general solution of Eq.(6], then by substituting this function in Eq.(S) one finds a general solution of Eq.(l) as an inverse function x i>i{y,Pi(y,C)). If an inverse function y - y,(p,C) is the general solution of Eq.(6], then this solution together with Eq.(5) give a general solution of Eq.(l] in the parametric form y = f '(piC), x $i(y,(p,C),p), where p is a parameter. There is a similar case when a general solution of Eq.(6] has the parametric form y = y,(t,C), p = pi[t, C), where f is a parameter. If an implicit relation fi(y,p, C) = 0 is a general solution of Eq.(6), then this function together with Eq.(5) give a general solution of Eq.(l) as an implicit parametric relation x = 0f(S,p), fi{y,p,C) 0, where p is a. parameter.

Notice that the Lagrange's and Clairaut's equations can be written as

and

y respectively, that is, as particular cases of Eq.(5).

Example 8. Find the general solution of the differential equation

y'3 - y - x = o.

Proceeding as stated above we obtain

x = y'3 - y = p 3 - y

1.5. NONLINEAR EQUATIONS 59

and

Further by integrating wefindy(p) as well as the general solution of the given equation in the parametric form

f =


where k is a number of solutions of Eq.(l), are some functions from the functions Pi. A solution of Eq.( 11 of the form y j in the conditions of tangency of two curves

1.5- NONLINEAR EQUATIONS 61

a) A singular solution may exist in some other cases. They can be found if a general solution of Eq.(l) is written as the implicit relation C) = 0. In this case a singular solution is the envelope of the single family of curves $(x, y, C) 0. First by eliminating C from the system

$ ' c ( I , y , C ) = 0, | y '

we obtain y = fii(x)i ' = 1, 2t . . . , 1 , where f^(x) are functions of x alone Further, proceeding as stated above we find all singular solutions of Eq.(l).

Example 12. Find the singular solution of the differential equation

y = x + y' + In j)'

if y = C + exp(i - C) is its general solution. According to Eq.(9) we obtain

y - C - e x p f i -C) = 0 , - l + exp(-r -C) = 0.

Eliminating C then yields y x + I . It is easy to verify that this function satisfies the given equation. Now we write the system (8)

x + 1 = C + exp(i - C), 1 = e x p ( i - C ) .

This system is compatible and it has the solution x C. Hence y == x + 1 is the singular solution of the given equation.

Complex solutions of E q . ( l )

In this case we must investigate Eq.(l) by the methods of analytic theory of differential equations. This problem is beyond the scope of our text.

Example 13. Find the solution of the differential equation

Proceeding as in Ex. 1 we obtain

y\(x)^e"'3(x-yf3, y2{x) = ?*(x-y)*'\ y'3(x) = e * " V - y ) 2 / 3

There exists on the real axis only solution yi(x). A general solution of the given equation exists in the complex plane.


1.5.3 Unsolvable Equations In order to solve these equations we must use some suitable methods. There does not exist a general method for solving of these equations

Example 14. Find the solution of the differential equation

x + y + y' + exp (x + y + y') 0.

Let or = x y y' is so real number that a = c~" It is obvious that 0 < a < 1. Now we obtain the linear nonhomogeneous equation

y' + y = -x - a.

Further proceeding as has been stated in Sec.1.4 we find the general solution of this equation

y = Ce" + 1 - x - a,

where G is an arbitrary constant. It is also a general solution of the given equation. Using Eq.(9) for this general solution, we convince ourself that the envelope of this single family of curves does not exist.

Notice that a general parametric representation of Eq.(l) can be written in the form

x = ip(u,v), JI = ^'(U,IJ), p = i?(u,i/),

where y , (6, t* are such functions of u and v alone, that Eq. (1) is satisfied identically. Then, taking into account that dy = y'dx and putting v v{u), we obtain

dv = tV - ' du ~ \b' - tV'

that is, the first order differential equation which is solved for the derivative. However, this equation is inlegrable by quadratures only in the case 1, as has been shown above, and it is not integrable by quadratures in the case 2.

PROBLEMS Find the solutions of the following equations, first solving them for y'(x). !. y-'-y^O. 2. y'1-4x:> = 0. 3. t / ' 2 - 4 j , 3 ( l - j , ) = 0. 4. ! / ' 2 - 4 | i , | = n. 5. y ' 3 + y2 -yy'W + 1) = 0. 6. 8y ' 3 -27y = 0.

7- y'3 ~ 4x ~ 0

8. yy'3 + x - l = 0. 9. t f ^ - V = 0. 10. F ' ^ + I ) - 1 = 0 .

1.5. NONLINEAR EQUATIONS

11. v V 2 + 1 - xy' = 0. 12. y " - 2 y y ' - j / 2 ( e 1 - l ) = 013. y J ) ' I - ( z y + l ) ! , ' + i = 0. 14. yy'{yy'-2x) - x 2 + 2y2 = Q. 15. y V * - 2xyy' + 2y2 - x2 = 0. 16. y* + y ? - l = 0. 17. y ' 2 -2xy' - Sx2 = 0. 18. y ' 2 - y 3 + y 2 = 0. 19. {xy' + Zy? -lx = Q. 20. y ' 3 - xy'2 - iyy' + 4iy = 0. Find the solutions of the following equations, first solving them for y(x). 21. 8 y ' 3 - 2 7 y = 0. 22. 2y' 3 + y ' 3 - y = 0.

23. 9"-Xy'-y+lxa=D, 24. y'2-ye' = 0. 25. y' 2 - e + 1 = 0. 26. xy'2 -xyy' - I = 0. 27. y'sin y' + cos y' y = 0.

28. r * + j s y - * r - t . 29. y' 2 + (x + l ) y ' - y = 0. 30. x 2 y ' 3 - i y ' + j , = 0. Find the solutions of the following equations, first solving them for x. 31. 1 = 0. 32. y' 3 + y ' - * = 0. 33. In y' + siny' x 0. 34. y' 3 - 4xyy' + 8y 3 = 0. 35. l y ' 1 + xy' - y = 0. 36. i y ' 2 - 2 y ' - z = 0. 37. y " + x y V - w 3 ' n y = 0. 38. x y ^ - y y ^ + l - O . 39. siny' + y' - x = 0. 40. y' s + ^ y ' + e 2 " = 0. Solve each of the following Clairaut's equations. 41. y = i y ' - y ' - 2 . 42. y = xy' + y,2 + x. 43. y = xy' + 2\/y' 2 + l -44. y = i y ' + j y ' 2

45. y = x y ' - y ' 2

46. y =ary' + y ' 3

Solve each of the following Lagrange's equations.


47. y = 3 x y ' - y ' 3 -48. y = 2 x y ' - y ' 2 . 49. y-2xy' - 4y' 3. 50. y = -xy' + y'2

51. S = ^ - 2 i ^ .

52. y = i x y ' + 2 ^ .

Find the singular solutions of the following differential equations using the given general integrals.

53. y' 2 + y 2 = 1, y - 2 s i n ( | + c) = 0 .

54. j r V + (2* " V)W + y2 = 0- xy - Cy + C ! = 0. 55. y ' 3 - 4xyy' + 8y2 = 0, y - C(x - C ) ! = 0

56. y' 2 - xy' - y + y = 0, - y + j + Cx + C 2 = 0.

Find the solutions of the following differential equations. 57. y ' ! - e '+ x = 0. 58. 8y' 3 - 12y'2 - 27(y - x) = 0. 59. x y ' - l n y ' + y = 0. 60. 2 y ' - l n y ' - x = 0. 61. x 2 y ' s - xyy' 3 - xyyn + y ! = 0. 62. x2y'2 - 2xyy' - 3y2 - x 2 = 0.

64. 3 j f M - p ' - y = 0. 65. y' 2 + 2xy' + y - 0 . 66. 2xy' - siny' - y = 0.

1.6 Applications in physics Here we consider some simple examples of how differential equations may arise in the analysis of problems in physics.

1.6.1 Mechanics

The second Newton's law can be written in the form

malt) = F ( r , v , f ) , ( l j

where m is the constant mass of a particle, r(Z) is its space coordinate at time t, v(t) = r(f) is its velocity at time (, a(t) v(t) - r ( i ) is its acceleration at time !,

1.6. APPLICATIONS IN PHYSICS 65

and F is an external force action on the particle with coordinates r and velocity v at time f. If the initial conditions

r(to) = m, r ( t 0 ) = v 0 (2)

are prescribed, then coordinates r(() are those solution of Eq.(l), which satisfy the initial conditions (2) at ( = t0.

Example 1. A body of mass m moves along of a straight line with no force upon it except a resistance proportional to the speed, that is, F = -kv, where k is a constant. Determine the total path of body up to stop, if its initial coordinate is 2ero and initial velocity is VQ > 0.

According to Eq.(l) we have

mi -- kx.

Upon integrating both sides of this equation we obtain

m i = kx + C i ,

where C\ is an arbitrary constant. Dividing of this equation by the term on the right hand side and integrating then yields

- ^ I n | - f c p + C i | = t + C7,

where Cj is an arbitrary constant. Hence

To determine the constants Ci and C2 we substitute this solution into the prescribed initial conditions, obtaining

0 = C, -exp(


Example 2. A body of mass m falls from rest under the gravitational force which is given by Newton's law, that is,

where G is the gravitational constant and i is the distance from the origin to the position of body at time f. Find the relation between the distance x and the time t. Determine the falling time of body on the origin from an initial distance H.

According to Eq.( 1) we have Gm

Multiplying the equation by i and integrating the obtained product gives, after some simplification,

* - G + C

where C\ is an arbitrary constant. Further we find

The negative sign is required because z decreases when 1 increases. The initial con-dition u(0) 0 requires that Ci = -G/H. Dividing this equation by the term on the right hand side and then integrating by making change of variable J : = f/smJs, we obtain

where C 2 is an arbitrary constant. The initial condition z{0) - H requires that

*H [W

Further we find, recalling that w/2 arcsin x = arccosi,

z arccos \l

H

where z > 0. Setting x = 0 gives the falling lime of body on the origin

T-*HJE-2 V2C7


1.6.2 Hydrodynamics

It is known that the viscous incompressible liquid flows from an orifice in the bottom of a vessel with the speed

u(ft) = ky/2gh(t), (3)

where k is a constant, g is the acceleration of gravity, k(t) is called the head of the liquid at time t. The continuity equation takes the form in these conditions

v(k)S(h) = const, (4)

where S(k) is the area of the upper surface of the liquid when the head is equal ft. Eq-s (3] and (4] and expression for the speed v(h) = ft make a system equations of motion of viscous incompressible liquid.

Example 3. Find the time required to empty a cylinder vessel full of water through an orifice of radius r in the bottom. The radius of the cylinder is i f and the height is H.

Taking into account Eq-s (3) and (4) and recalling that S{h) nR2 and {0) 7rr!, we obtain

v{h) = k^2g~k.

The initial condition is h(0) = H. Hence

Dividing this equation by ft"2 and integrating both sides gives, after some minor calculations,

where Ci is an arbitrary constant. The initial condition requires that C, = VH. Hence

The time required to empty the vessel is determined by the condition h(T) = 0. This condition then yields


1.6.3 Electrical Networks

According to Kirchhoff's second law: In a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of circuit, that is,

U1+Ui+...+Un=e, (5)

where U[ are the voltages drop across the conductors in the circuit, is the impressed voltage. The voltages drop across a resistance, a capacitor and an inductance are equal to

respectively, where I(t) is the current at time i , R is the resistance, C is the capaci-tance, L is the inductance and Q is the total charge on the capacitor.

Example 4. Determine the temporal variation of the current in a series circuit consisting of a resistance R, a capacitance G and an impressed voltage after the switch is closed. Find the current at the initial time.

According to Eq-s (5) and (6) we have

RI(t) + Qp- = , at O f t

The initial condition is Q(0) = 0. Since 7 = Q we obtain

RQ(t) = E-

Dividing this equation by the term on the right hand side and integrating both sides give

- H C l n |f-^| = f + Ci,

where Ci is an arbitrary constant. The initial condition Q(0) 0 requires that

C, = -RC]n\\.

Putting > 0 we obtain Q(t) = C ( l - e - " f l c ) .

and

The current at ( = 0 is 1(d) jR. Example 5. Determine the temporal variation of the current In a closed series

circuit consisting of a capacitor C and an inductance L if at r = 0 the initial current is la > 0 and the initial charge on the capacitor is zero.



The initial conditions are 1(0) - I 0 and Q(0) = 0. Since I(t) = Q(t) we obtain

In a way analogous to that used in Ex.2 we find

where C, is an arbitrary constant. Hence

Q

The initial conditions require that sign is positive and C, 1$. Then dividing this equation by the term in the right hand side and integrating both sides gives

v'LCarcsin f%= | = f + t?3, \l0sfw)

where Ci is an arbitrary constant. Hence

Q(t) - loVLCsin

and

The initial conditions require that Cj 0. Hence

1(1) = IQCOS

I.6.4 Kinetic Theory

1. It is known that liquid evaporates in the air with the rate

dM(i) dt

= kS(p(t)-Ps), (7}

where M( t ) is mass of the liquid at time (,jfe is a constant, S is the area of the evaporating surface, p(f) is the vapor pressure at time i , p, is the saturated vapor


pressure at constant temperature. The Eq.(7) Is correct when the vapor distributes spatially uniform into the surrounding air, that is,

where po is the vapor pressure at the initial time ( 0 . A is a constant, m(t) M$-M{t} is the mass of the vapor al Ihe Lime '. Mo is the mass of the liquid at the initial time fa , V is the volume of the surrounding air.

Example 6. Determine the temporal variation of the mass of water during its evaporation from a cylinder vessel of volume Ua, if u 0 < < V p(0) = p 0 , M(0) = Mo.


where v{t) is the volume of water at time t, p is the density of water. Since V >> "o > "( ' ) then neglecting of u(I) in the first equation gives

= k S ( A v + p - p '

Integration of this equation (see Ex.4) then yields

. I M - M(t) kASt n I " p i ~ L + P*-P> = - - y - + C

where C is an arbitrary constant. By the initial conditions we find C = In |/>o - p,| Putting p, > po we obtain

M ( 0 = M o - ^ ( p , - p D ) ( l - e - M S " 1 ' ) .

2. It is known that the pressure of the gas at an altitude z can be determined as

p(z) = j p(z)g(z)dz, (9)

where p(z) is the density of gas at the altitude z, g(z) is the acceleration of gravity at this altitude. Then the Clapeyron ideal gas law

< W - 3 ^ . ( . . )

where R is the gas constant, T{z) is the temperature of gas at the altitude z, together with Eq.(9) make a system of equations for p, p and T.


Example 7. Assuming that the gravitational force is constant, determine the behavior of the pressure of the isothermal ideal gas. The gas pressure at z 0 is equal r^.


p(z)g dz RT '

that is, an integral equation for p(z). After differentiation with respect to z we find

dp = pg dz RT'

Integrating (see Ex. 4) then yields

\np(z) = - ^ + C,

where C is an arbitrary constant. Notice that p is intrinsically positive so that absolute value is not needed. The initial condition gives C Inpo. Thus we obtain

P = P o e - " " t r ,

so called Boltzman's formula. 3. According to Newton's law of cooling, the cooling rate of a body is given by

the equation

^ . = -h(T(t)-Tm), (11)

where k is a constant, T(t) is the temperature of the body at time t, TCIt is the temperature of the surrounding medium.

Example 8. Determine the temporal variation of temperature of a body into the surrounding medium with the constant temperature T , , if the initial temperature of the body is Zjj.

According to Eq . ( l l ) we have

Integrating (see Ex. 4) leads to

In \T(t)-T0\ = -kt + C,

where C is an arbitrary constant. The initial condition requires that

C = bi ITo - IW, ! -

Putting T0 > Ten w e obtain


1.6.5 Nuclear Physics

It is known that the rate of the radioactive decay is given by the equation

dm{t) dt = - M O .

where k is a decay constant, m(f) is mass of radioactive material at time (. Example 9. Determine the decay constant of Radii if its half decay time is 1600

years. According to Eq.(12) we have

m'(f) = - M O '

An integration then yields In m(t) = -kt + C,

where C is an arbitrary constant. Notice that m is intrinsically positive so that absolute value is not needed. The initial condition m(0) m 0 requires that C -In mo. Then

m(t) m n e x p ( - t t ) ,

where m 0 is the mass of Radii at the initial time. The half decay time T,/j is deter-mined by the condition

Hence we obtain k = {]n2)/Tl/2 = (In2)/I600 (measured in 1/year).

16.6 Optics

It is known that the intensity of light into absorbing medium at the point z + Sz is determined as

I(z + Sz) = I(z)-kl{z)Sz,

where / ( ; ) Is the intensity of light at the point z, k is the constant of absorption. Subtracting ! { : } , dividing by Sz and approaching the limit 6z 0 leads to the equation for the intensity of light

9 " * * * (13) which is completely analogous to Eq.(12).

PROBLEMS 1. A body of mass m falls from rest in air under a constant gravitational force

and a resistance proportional to each of the following powers of the speed: a) first; b) second Determine the falling time of body in each case if body's initial altitude is equal to H.


2. Suppose everything is as in Problem 1, except that resistance is Fr = k,v k2v2 Determine the falling time of body. Find the limiting velocity in each of the following cases: &] i 0; k^ 0.

3. A body of mass m is thrown vertically upward from the surface of the earth under a constant gravitational force and a resistance proportional to each of the following powers of the speed : a) first; b) second. Determine the maximum height attained by the body, and the time at which the maximum height is reached, if bodies Initial speed is u 0 .

4. Suppose everything is as in Problem 3. Determine the time at which the body returns to its starting point. Approach each of the following limits: the resistance is very large; the resistance is very small.

5. A body of mass m moves along of a straight line under a resistance force proportional to (1 + o)-th power of the speed. Determine the temporal variation of the coordinate and velocity if body's initial speed and coordinate are vu and zero, respectively. Approach each of the following limits: a 0.5; a < 0; a 1; a > 2.

6. A body of mass m moves along of a straight line under the Hook's force F kx. Determine the temporal variation of its coordinate, if at ( = 0 the initial speed is zero and the initial coordinate is x0.

7. A body of mass m falls from the rest under a force F k/x3, where k is a constant and x is the distance from the origin to the position of body at time t. Determine the temporal variation of bodies coordinate, if the initial distance is h.

8. Find the time required to empty each of the following vertical vessels full of water: a) the cone of height H and top base of radius R, b) the cone of height H and bottom of radius R\ c) the sphere of radius R; d] the circular paraboloid z = (z 3 + y 3 ) / / / , 0 < z < H; e) the circular paraboloid 4 = H - {x2 + y2)/H, 0 < 2 < H. Assume that the radius of an orifice in the bottom is equal to r. Note that the spherical vessel has also a small hole at the top.

9. Find the time required to empty a horizontal cylinder full of water through an orifice of radius r in the lower point of cylinder. The radius and the length of the cylinder are R and H, respectively.

10. Determine the temporal variation of the current in a series circuit consisting of an inductance , a resistance R and an impressed voltage , if at ! = 0 the switch is closed.

11. Determine the temporal variation of the current in a series circuit consisting of a capacitance C, a resistance R, if at t = 0 the capacitor has a total charge equal to Q, and the switch is closed.

12. Determine the temporal variation of the current in a dosed series circuit consisting of an inductance L , a resistance R, if at t = 0 the current is equal to la

13. Determine the temporal variation of the current in a closed series circuit consisting of an inductance L and a capacitance C, if at ( = 0 the capacitor has the maximum total charge Q 0 > 0.

14. Determine the temporal variation of the mass of water which evaporates from


a full cylinder vessel of the volume va inside another closed vessel of the volume V > v Suppose that at t = 0 the mass of water is equal to m 0 , the vapor pressure p 0 < p. Approach the limit as (p, - p 0 ] ^ 0 at V va-

15. Determine the temporal variation of the mass of water which evaporates from the full cone vessel of a height H and a radius of the top base R Assume that urj < < V and p(0) = pr, < p,

16. Determine the air pressure above earth level, taking into account that the acceleration of gravity on an altitude ft is g(k) = So/(l + k/R)2, where gB is the acceleration of gravity at ft = 0 and R is the radius of earth. Assume that the air is the ideal gas at the constant temperature T.

17. Find the differential equation for the pressure of the "ideal" isothermal self-gravitate gas into a spherically cloud.

18. The body in air at 20 C cools from 100 C to 60 C in 10 minutes. When will its temperature be 25 C?

19. Find the time required to evaporate a spherical liquid drop of a radius R. Assume that V R3 and p(0) = po < p,-

20. Determine the temporal variation of temperature of a body suppose that the temperature of the surrounding medium is a function of the time Ttn{t). Assume that 7/(0] = Te.

21. I t is known that 206 gram of Plumb is produced owing to the radioactive decay of 238 gram of Uran. Determine the age of such rock, where the relation of Uran mass to Plumb mass is equal to 50:7. The half decay time of Uran is equal to 4.5 x 109 years. Assume that at f = 0 the mass of Plumb is equal to zero.

22. A ray of light gets into a layer of water bounded by two parallel mirrors. Determine the profile of temperature into the layer of water, if the distance between mirrors is d, the intensity of the light is I 0 , the duration of expose is T and the coefficient of absorption is fc. Assume that the ray of light is normal to the surface of mirrors.

23. What's the share of light that has been absorbed into the layer of water thickness of 2 m, if half of the light has been absorbed into the layer of thickness 0.35 m.

1.7 Miscellaneous problems This section consists of a list of problems, which can be solved by the methods of the previous sections. They are presented so that the reader may have some practice in identifying the methods applicable to a given equation

PROBLEMS 1. (6x + y - l)dx + [ix + y - 2)dy = 0. 2. y (iV + 1) dx +

1.7. MISCELLANEOUS PROBLEMS 75

5. y3dx + 2 (x 2 -xy*)dy = Q. 6. *yd#+'fyt -x6)dy = 0. 7. (x - 2y + l)dy - (2* - n + l)dx = 0. 8. (8z + 2 5 y - 62)dx- ( l l x + 4 y - U)dy = 0. 9. (x + 2y + l)dx - (2x + 4y + 3)dy = 0. 10. 4xy 2dx + (3x 2y - 1) dy = 0. 11. (x + y - 2)dx + (x - y + 4)dy = 0. 12. 2 ( v W + 1 - x2y) dx - s^ rfs - 0. 13. (2is iny - y2 sinx) dx + (x2 cosy + 2ycosx + 1) dy = 0. 14. ( i l n y - x2 + cosy) dy + (x3 + y Iny - y - 2xy) dx 0. 15. (x 2 - 4xy - 2y 2) dx + (y 2 - 4xy - 2x2) dy = 0. 16. (6sy + x1 + 3) jf* + 3y2 + 2xy + 2s - 0. 17. (2xcosy - y 2 sin x) dx + (2y cos I x 2 sin y)dy 0.

xdx + ydy ydy - xrfy _ 71 + x* + y> x2 + y2

19. (2y + xy1)dx + (x + x2y2) dy = 0. 20. (1 + x2y)dx + x2(x + y)dy = 0. 21. (x J + y 2 + 2x) dx + 2ydy = 0. 22. (x 2 + x2y + 2xy -y2 -y3)dx + (y 2 + xy2 + 2xy - x2 - x 3 ) dy = 0. 23. {2x3 + 3z 2y + y2-y3)dx + (2y3 + 3zy ! + x2 - x3) dy = 0. 24. y 2 r i i + ( i y - l ) d y = 0. 25. (2x 2y + x)dy + (y + 2xy2 - x2y3) dx 0. 26. (z 3 + xy2 -y)dx + (y 3 + x2y + x) dy = 0. 27. (x 2 + y)dy + x ( l - y)dx = 0. 28. (x2y3 + y)dx + (x3y2-x)dy = Q. 29. 2y 3y' + xy 2 - x 3 = 0. 30. y' sec 2y + x tan y = x. 31. e-'y' - e"1 = e. 3 2 y = _ 2 y i T | =

v T + y i

V l + y 2 34. 3dy + ( l + e I + 3 v)dx = 0.

35. 33/ - ysinx + 3y4 sinx = 0 sinz

36. y + sin x tan y . cosy

37. y'cosx - ysinx = y* 38. zy" + y = x y 2 l n x . a o . y ' + ^ y ' U - x 2 ) , y(I) = l .

X

40. y 2 (x + a)dx + X (x2 - ay) dy = 0. 41. (2xy - i ' y - y3) dx - (x 2 + y2 + x3 - xy2) dy = 0. 42. (3x'y 2 + y 5 ) dx - (xy 4 + 2x 5y) dy = 0.


43. (x2 + 2xy + x2y -y1-y1)dx + (y2 + xy2 + 2xy - x 2 - x 3 ] dy = 0. 44. * V - x 2 y 2 + 5xy - 3 = 0. 45. i V - i / ' - x 2 y + x1 = 0. 46. y ' + V + ^ - - 2 = 0.

47. y' = a , 2 - x 2 + l . 48. (x - x

C h a p t e r 2

N - t h O R D E R D I F F E R E N T I A L E Q U A T I O N S

2.1 Reduction of order The order of an n-th order differential equation can

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