GIFT F
the estate of
Professor William F. Mever
A SHORT COURSE ON
DIFFERENTIAL EQUATIONS
A SHORT COURSE ON
DIFFERENTIAL EQUATIONS
BY
DONALD FRANCIS CAMPBELL, PH.D.
PROFESSOR OF MATHEMATICS, ARMOUR INSTITUTE OF TECHNOLOGY
THE MACMILLAN COMPANY
LONDON: MACM1LLAN & CO., LTD.
1907
All rights reserved
ASTRONOMY LIBRARY
COPYRIGHT, 1906
BY THE MACMILLAN COMPANY
Set up and electrotyped. Published September, 1906
Enlarged October, 1907.
PRESS OF
THE NEW ERA PRINTING COMPANY
LANCASTER, PA.
moi
PREFACE
LIBRARY
In many Colleges of Engineering, the need is felt for a text
book on Differential Equations, limited in scope yet comprehen
sive enough to furnish the student of engineering with sufficient
information to enable him to deal intelligently with any differen
tial equation which he is likely to encounter. To meet this need
is the object of this book.
Throughout the book, I have endeavored to confine myself
strictly to those principles which are of interest to the student of
engineering. In the selection of problems, the aim was con
stantly before me to choose only those that illustrate differential
equations or mathematical principles which the engineer may
meet in the practice of his profession.
I have consulted freely the Treatises on Differential Equations
of Boole, Forsyth, Johnson, and Murray. I am indebted to two
of my colleagues, Professors N. C. Riggs and C. W. Leigh, for
reading parts of the manuscript and verifying many of the
answers to problems.
D. F. CAMPBELL.
CHICAGO, ILL.,
September, 1906.
PREFACE TO ENLARGED EDITION
This book as it first appeared consisted of the first eight
chapters as here given. The kindly criticism by a number of
:hose teachers for whose use it was intended on the need of a
liscussion of equations, that occur in investigations in Mathe
matical Physics, other than those given in these chapters has
nduced me to add Chapter IX to the book.
M577086
vi PREFACE
In the preparation of Chapter IX. , I have drawn freely from
Professor Byerly s Treatise on Fourier s Series and Spherical
Harmonics, from Professor Bocher s pamphlet entitled Regular
Points of Linear Differential Equations of the Second Order and
from notes kindly loaned me by Professor Snyder of Cornell
University. I have also consulted Heffter s Treatise on Linear
Differential Equations with one Independent Variable.
To those teachers who have sent me their criticism of the
book in its original form, as well as to others who have cordially
received it, I am under the deepest obligations.
D. F. CAMPBELL.
CHICAGO, ILL.,
June, 1907.
CONTENTS
PAGES
CHAPTEE I
INTRODUCTION 1-15
CHAPTEK II
CHANGE OF VARIABLE 16-21
CHAPTEK III
ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND
FIRST DEGREE 22-40
CHAPTEK IV
ORDINARY LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT
COEFFICIENTS . 41-64
CHAPTER V
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS. EXACT
LINEAR DIFFERENTIAL EQUATIONS 65-70
CHAPTER VI
CERTAIN PARTICULAR FORMS OF EQUATIONS .... 71-76
CHAPTER VII
ORDINARY DIFFERENTIAL EQUATIONS IN Two DEPENDENT
VARIABLES 77-88
CHAPTER VIII
PARTIAL DIFFERENTIAL EQUATIONS 89-96
CHAPTER IX
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS. INTE
GRATION IN SERIES . 97-123
Tii
A SHORT COURSE
ON
DIFFERENTIAL EQUATIONS
CHAPTER I
INTRODUCTION
1. There are various definitions given for a function of one
variable. We shall here adopt the following :
If to every value of x there corresponds one or more values of
/(#), then f(x) is said to be a function of x.
This definition includes a constant as a function of x, for if
/(#) is constant, then for every value of x, /(#) has a value,
namely, this constant.
A definition of a function of two variables is the following :
If to every pair of values of two variables x and y there cor
responds one or more values of /(a, y), then /(a?, y) is said to be
a function of x and y.
This includes a constant or a function of one variable as a
function of x and y.
A function /(#) of one variable x is single valued when for
every value of x there is one and only one corresponding value
A function f(x) of one variable x is continuous for a value
x = a if /(a) is finite, and
[/( + h)} = 1" [/(a - A)] = /(a). ; ...
A function /(#, y) of two independent variables x and y is
single valued when for every set of values for x and y there is
one and only one corresponding value off(x, y).
I
2 SHORT COURSE ON DIFFERENTIAL EQUATIONS
A function f(x, y) of two independent variables x and y is
continuous for a set of values x = a, y = 6 if /(a, 6) is finite,
and
limit |~ ~|
no matter how h and & approach zero.
The following definitions are given in almost any work in
calculus :
If /(#) is a single valued and continuous function of x, given
by the equation y = /(#), then
Az and Ay denote the increments of x and y respectively,
dy^ _ limit
A~ Ax =
dy = -= d#.
cfo
If /(#) is single valued and continuous, and dyjdx is contin
uous, then
ft l/4\
<fo 2 dx\dx/
In general, if /() is single valued and continuous, and the
preceding derivatives are all continuous, then
._
dx n ~ dx\dx
If f(x, y) is a single valued and continuous function of two
independent variables x and y, given by the equation z =/(>, y),
then dz/dz is the derivative of 2 with respect to x when y is held
constant ; dz/dy is the derivative of z with respect to y when a is
held constant.
2. In a single valued and continuous function f(x) of one vari
able x, given by the equation y =/(#), whether x is the inde
pendent variable or a function of some other variable or variables,
we have
INTRODUCTION
= d(dx) ; d*x
Definitions. The differentials dx, d\ d*x, , d n x, or
dy y d*y, d 3 y, - , d n y are called the first, second, third, , nth,
differentials respectively.
3. Derivation of d*y and cfy when no assumption is made re
garding x being independent or a function of some variable or
variables.
By taking differentials in succession any differential may ulti
mately be found.
4. In the differentials of the preceding article, if x is an inde
pendent variable, it can be assumed without loss of generality,
that A#, or what is the same in this case, dx, is constant. That
is, it can be assumed that x changes by equal increments. Under
this supposition, therefore, d*x and all higher differentials of x
can be taken zero. Therefore, under this supposition,
The place which a derivative or differential occupies in the
succession of derivatives or differentials indicates the order of the
derivative or differential. Thus, a second derivative or differ
ential is said to be of the second order, a third of the third order,
and so on.
4 SHORT COURSE ON DIFFERENTIAL EQUATIONS
5. The only functions usually considered in elementary works
in calculus are functions of a real variable. Such functions with
one exception are the only ones considered in the following pages.
The exception is e* where z is a complex quantity.
The student is already familiar with the definition of e x where
x is real. He is, however, probably not familiar with the defini
tion of e? when z is a complex quantity. A definition of this
function will now be given.
The infinite series
where z is a complex quantity, can be shown to have a determi
nate, finite value for every value of z. It also reduces to the
infinite series
when z becomes real and equal to x, and this series, it will be re
membered, is equal to e x for all values of x. It therefore appears
that the infinite series in z would be satisfactory as a definition
of e z . We shall define e z by saying that it is equal to the infinite
series
for all values of z.
From this definition, the following theorem can be established :
Theorem. If z = x-\-yj where x and y are real, and j = V 1>
then
e* = e*(cosy -f jsiny).
Proof.
2 gS
e* = l-fz + ur + r^ + ---,ky definition.
N , O + J0 ) f , O + ^7 ,
INTRODUCTION 5
Consider all the terms containing x r . These are found from the
terms
O + wY , (^wY^
ll lifl
They are
Fi ,,,, M! , (JL)!^1_
II
or
af \^ -.(> .()*. (s/) 4 . (Y. ..1
7[i + y> + - -g-- -JT -[5- J-
Separate the real terms from the imaginary and there results,
or
Let r take all positive integral values in succession from 0.
In this way we get all the terms of the development e z . Then
e* = 1 1 + re + j|- + r| + -J [cosy +./siny]
The theorem is therefore proved.
EXAMPLES. e~ x+ "^ = e~ x (cos Bx + j sin 3a;)
EXERCISES
1. Given y = logo;, find %, rf 2 ^, c? 3 ?/ :
(a), on the assumption that x is the independent variable ;
(6), making no assumption with regard to x.
In the results of (6), substitute x = cos and show that the
results are the same as those obtained by first substituting the
value of x in log x and then taking the differentials.
2. Given y = e x where x cos 0, express dy, d?y, d?y in terms
of without substituting the value of x in the equation y = e?.
6 SHORT COURSE ON DIFFERENTIAL EQUATIONS
3. Given y = log x where x = sin 6, express dy, d*y, d*y in
terms of 6 without substituting the value of x in the equation
y = log x.
4. Prove that & ** = e z (cos y j sin y~).
5. Prove that e^e"*** = e x+z+(y+w)J , where x, y, z and w are
real.
ANSWERS
i / N j dx ,. efce* ,, 2cta 8
l(a). <fy = ; d 2 </= -^-; ^ = ^3-. .
3xdxd*x -\-2dx*
2. dy = smO e cos *dO ;
^ 2 y = _ sin0 e cos ^ 2 ^ -|- (sin 2 ^
d?y = smB e cose d 3 + 3(sin 2 ^
+ sin0(l
3. rfy = cot0 d0 ; ePy = cot0 <f0 _ cosec 2 rf0 2 ;
<P = cot0 c? 3 _ 3 cosec 2 c?0cZ 2 2 cosec 2 cot0 d0*.
6. Definition of differential equation. A differential equation
is an equation involving derivatives or differentials with or with
out the variables from which these derivatives or differentials are
derived.
The following are examples of differential equations :
(3)
INTRODUCTION
7. In examples (1) to (4) inclusive of the preceding article
it will be noticed that differentials enter the equation only in de
rivatives. It is conceivable, however, that there might be an
equation containing differentials other than those in the deriva
tives, as for example,
but there is no need of entering into a discussion of such equa
tions, and we shall not do so. In what follows, we shall assume
that if the equation is written in differential form, the differen
tials can all be converted into derivatives by the process of
division.
8. Classes of differential equations. Differential equations
are divided into two classes : ordinary and partial.
An ordinary differential equation is one in which all the
derivatives involved have reference to a single independent
variable.
A partial differential equation is one which contains partial
derivatives and therefore indicates the existence of two or more
independent variables with respect to which these derivatives
have been formed.
Thus, in Art. 6, equations (1), (2), (3) and (4) are ordi
nary differential equations, and equations (5) and (6) are par
tial differential equations.
Chapters I to VII inclusive are devoted to a discussion of ordi
nary differential equations. Chapter VIII contains a short
treatment of some partial differential equations.
9. Order and degree of a differential equation. The order of
a differential equation is that of the highest derivative or differ
ential in the equation.
8 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Thus, in Art. 6, equations (1) and (4) are of the third order,
and (2) and (3) of the second order.
The degree of a differential equation is the degree of the deriv
ative or differential of highest order in the equation after the
equation is freed from radicals and fractions in its derivatives.
Thus, in Art. 6, equation (1) is of the second degree, equa
tions (2), (3) and (4) of the first degree.
10. Solutions of a differential equation. Let us consider the
differential equation in each of the two following examples, and
see if, from the equation, we can get a relation connecting x and
y and not involving derivatives, such that, if the value of y in
terms of x be substituted in the equation, the equation is satisfied.
EXAMPLE 1. -/- x*.
dx
By integration, we get
a?
EXAMPLE 2. -f y = 0.
Multiply the equation by 2dy/dx and integrate.
. . y = A/^sin (x -f c x ), or y = cos (x -f c 2 ).
In example 1, if J# 3 + c be substituted for y in the equation,
there results x* = x*. The equation is therefore satisfied.
In example 2, if Vc sin (x + c,), or =h Vc cos (x -f c 2 ) be
substituted for y in the equation, there results, in the first case,
INTRODUCTION 9
zp Vc sin (x -f Cj) Vc sin (# -f c x ) = 0, and in the second case,
q= Ve cos (# -f c 2 ) A/C cos (z -f c 2 ) = 0. In either case the
equation is satisfied.
Definition. A solution of a differential equation is a relation
between the variables of the equation and not involving deriva
tives, such that if the value of the dependent variable be substi
tuted in the equation, the equation is satisfied.
Thus, y = J# 3 -f- c of example 1, and y = Vc sin (x -f cj
of example 2, are solutions of the equations.
In this book we shall not concern ourselves with the question
of whether every differential equation has a solution but shall be
content with finding solutions in the few special cases discussed
here.
11. A solution of an ordinary differential equation may be
one of three kinds : general, particular and singular.
A general solution is one which contains arbitrary constants
equal in number to the exponent of the order of the equation.
Thus, in example 1, Art. 10, the number of arbitrary con
stants is one and the exponent of the order of the equation is 1,
and in example 2 of the same article the number of arbitrary
constants is two, and the exponent of the order of the equation
is 2. In either case the solution is the general solution of the
equation.
A particular solution of a differential equation is a solution
obtained from the general solution by giving one or more of the
constants particular values.
Thus
3? X* 3?
2/= 3 2/ = "3 + 1 r y = 3 ~ 5
of example 1, Art. 10, or y = sin x, y = 2 sin x, or y= 3 cos x,
of example 2 of the same article, are particular solutions of the
equations.
A singular solution of a differential equation is a solution with
out arbitrary constants which cannot be derived from the general
solution by giving the constants particular values.
2
10 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Singular solutions will not be considered in this book.
12. A solution of a differential equation is not a general solu
tion unless the constants are in number equal to the exponent of
the order of the equation, and cannot be reduced to a fewer
number of equivalent constants.
Thus, y = ce x+a , c and a arbitrary constants, although it con
tains two arbitrary constants, is not the general solution of a dif
ferential equation of the second order, as can readily be shown.
The equation y = ce x+a is the same as y = c a e x . Now c a is equiv
alent to only one arbitrary constant because an arbitrary con
stant can have any value and thus all the particular solutions
got by giving c and a all possible values can be obtained. There
fore y = e a e x is equivalent to a solution y Ae x , A arbitrary, and
cannot therefore be the general solution of a differential equation
of the second order.
13. Let y =/ 1 (^), y =/ 2 O), , y = /() be solutions of a
differential equation.
Definition. If the c s cannot be chosen, not all zero, such that
c i/i(^) + ^/aOO + + C n/n(^) i identically zero, then the
solutions are said to be linearly independent.
Thus, y = d= Ve sin (x 4. c,) and y= =t Vc cos (x -)- c 2 ) of ex
ample 2, Art. 10, are such that no values c s and c 4 , not both
zero, can be chosen such that c s Vc sin (re-f-Cj) c 4 Vc cos (#-[-c 2 )
is identically zero. The solutions are therefore linearly inde
pendent.
14. Derivation of an ordinary differential equation. Let
*(x, y, Cl ) = (1)
be an equation containing x and y, and the arbitrary constant c r
By differentiation of (1) there results
j> * o . (2)
dx T cty dx
Equation (2) will in general contain c r If between (1) and
INTRODUCTION 11
(2), c l be eliminated, the result is a differential equation of the
first order of which <j>(x, y, Cj) = is the general solution.
EXAMPLE. Find the differential equation of which
m _ za
is the general solution.
dx ~ 1
Eliminate c t between the equations. Therefore
dy _
- 7 - -f- 2xy = mx
dx
is the differential equation of which
y = - + Cl <T"
is the general solution.
Sometimes the arbitrary constant is so involved that it disap
pears in the equation which results from the differentiation. In
such a case this equation is the desired equation.
EXAMPLE. Find the differential equation of which y 2 =
is the general solution.
Divide both sides of the equation by x.
:. - = 2c r
x
By differentiation there results
which is the desired differential equation.
Let
<f>(x, y, c v c 2 ) = (1)
be an equation between x and y, and two arbitrary constants c v
and c 2 .
12 SHORT COURSE ON DIFFERENTIAL EQUATIONS
By differentiation of (1) there results
=.
dx dy dx
Equation (2) contains dy/dx and will in general contain c x and
c 2 also. Eliminate one of the constants between the two equa
tions. Suppose the constant c l to be eliminated. The resulting
equation contains dy/dx and in general x } y and c 2 . Call it
By differentiation there results
dij, d^dy
dx + dydx= Q
Equation (3) contains d*y/dx* and will in general contain c 2 .
Eliminate c 2 between (2) and (3). The result is a differential
equation of which <j>(x, y, c,, c a ) = is the general solution.
EXAMPLE. Find the differential equation of which
is the general solution.
Differentiate y =
Eliminate c v
dy 2c.
. *. w x - - -.
dx x
Differentiate.
Eliminate c between
cPw 2c, , ^y 2c 2
y^ 2 = * and y x-~ = ?
dx* x* y dx x
INTRODUCTION 13
which is the desired differential equation.
15. It is seen from the preceding article that one constant can
be removed after each differentiation. From this it would be
expected that, starting with the differential equation, an arbi
trary constant might be introduced every time the order of the
differential equation was lowered by unity. Then, since lower
ing the order of a differential equation of the nth order by unity
n times would result in a solution of the differential equation, it
would be expected that a solution would contain not more than
n arbitrary constants.
It is a theorem that a differential equation cannot contain a
solution having more arbitrary constants than the exponent of
the order of the equation unless the constants are such that they
can be reduced to a fewer number of equivalent constants. This
will be assumed without further discussion.
It is also a theorem that a differential equation cannot have
more than one general solution. This theorem will be assumed
without discussion.
16. A genera] solution may have various forms but there is
always a relation between the constants of one form and those of
another. Thus, the general solution of example 2, Art. 10, may
be written y = A sin x -{- B cos x instead of y= Vc sin (z-f-Cj).
This latter form of solution is y = . Vc cos c l sin x V^sin C L cos x,
so that A = Vc cos c and B = V^sin c
j EXERCISES
1. Determine the order and degree of the following equations. ft Pv
14 SHORT COURSE ON DIFFERENTIAL EQUATIONS
In each of the seven following exercises determine the differ
ential equation of which the given equation is the general solu
tion, given that c lt c 2 and c are arbitrary constants.
2. 2/ = c i si* 1 mx + C 2 cos mx - 5- y = ex -f c c 3 .
3. v = e l cos (mt -j- c 2 ). y^ 6. #?/ = e^ -f- c 2 e~ z .
^ 4. O - c x ) 2 4. (y _ c 2 ) 2 = m 2 . 7. ^ - 2c.r _ c 2 = 0.
9. Show that
is a solution of
^ ,
x dx*
v 10. Show that
4y = 3^ + c/ -f
is a solution of
*^_ 54^ + 5
^ 2 dx T ^
11. Show that
y = T V" + W-* -f
is a solution of
3+ 4 2+^=
12; Show that
c,
v = r + C 2
is a solution of
<fv 2 <iv
ANSWERS
o ^ 2y 2
INTRODUCTION 15
4 i ,
CHAPTER II
CHANGE OF VARIABLE
17. Interchange of Variables. It is sometimes desirable to
transform an expression involving derivatives of the function y,
in y = /(#) where x is the independent variable, into an equiva
lent expression involving derivatives of the function x, given by
the same equation, where y is the independent variable.
The formulas for such a transformation can be readily estab
lished as follows:
limit
limit
d?y
dx*
,(*y\
dx\dx)
d / dy\ dy
dx
dy*
/dxV dx
(dy) dy
dx 3
d
dxdx
D-
tfx
W
dy
_dtf_
(dy)
(i)
dz dz dy
since -=- = -y- j-.
ax ay ax
by substitution from (1).
(2)
l_
dx
dy
16
CHANGE OF VARIABLE 17
dx\ 3 d?x itfx\ 2 /<M 2 <f * efo /d!Vv 2
) (dy) -dtfdy + 6 (dy*)
The method of procedure for higher derivatives is evident.
The transformations to which these formulas apply are called
change of the independent variable or interchange of variables.
EXAMPLE. Change the independent variable from x to y in
the equation
/*yV <*y*y ^W^/Y_o
\dx*) ~ dx dx* ~ dx* \dx) ~~
Substitute from (1), (2) and (3).
.-. 3
18. Change of the dependent variable. Suppose that y is a
function of x and at the same time is a function of some other
variable 2. The derivatives of y with respect to x can then be
expressed in terms of derivatives of 2 with respect to x.
As a function of 2, let y = <(z). Denote differentiation with
respect to 2 by primes. Then
= = = _
dx ~ dz dx dz dx dx
18 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Similarly for higher derivatives.
The above transformation is called change of the dependent
variable.
EXAMPLE. In the equation
change the dependent variable from y to z where y = tan z.
dy dz
f- = sec z-y-.
ax ax
,
~ = 2 sec 2 2 tan z I -7- ) + sec 2 -7-,.
<fo 2 2
Substitute in the equation.
4 z tan i +sec 4 z 2 -(2 tanz_l) Bec**
3a; sec 4 2 -7- = 0.
19. Change of the independent variable. Suppose that y is
a function of x where a is a function of some other variable z.
The derivatives of y with respect to x can then be expressed in
terms of derivatives of y with respect to z.
As a function of z let x = <(z). Denote differentiation with
respect to z by primes. Then
dy dydz dy 1 dy \
dx ~ dzdx~ dzdx ~ dz <#> (z)*
d*y d (dy\ d f dy 1 \ 1 1 d*y <"(z) dy
~dz
CHANGE OF VARIABLE 19
A ^_{^y\ A\ __ L_ d *y _ ~
{* (*) } 6 <fe-
Similarly for higher derivatives.
The above transformation is called change of the independent
variable.
EXAMPLE. In the equation
d*y x dy y _
"cfo -r^^+l-* 2 -
change the independent variable from x to 2, where x = cos z.
dy dy dz dy
-/- = -f - cosec z -f .
dx dz ax ciz
d y d(dy\ d(
= = -
dz
L dy d*y
= cosec 2 zQQtz~j-+ cosec 2 3 -j4.
ds dz 2
Substitute in the equation.
. . cosec 2 z -r| - cosec 2 z cot z^ + cosec 2 2 cot 2 ^+cosec 2 2 2/=0.
^*y A
. * . -TT + ?/ = 0.
d2 8
When changing either the dependent or independent variable
to a third variable, it is better to work out each derivative in the
particular case considered rather than use the derivatives ex
pressed in the general case as formulas.
20 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXERCISES
In each of the four following exercises, change the independent
variable from x to y.
f 1 d*y dy ^ d 2 y / dy\ z / dy\ 3
da?
d>yd y 3 ., o r
4 - 8 +8 ~ ~ =
In each of the two following exercises change the dependent
variable from y to z.
^.a+^
where = tan 2.
where ?/ = e z .
In each of the four following exercises, change the independent
variable from x to z.
7. x* -j-? 4- z / + w = 0, where a; = c*.
c?^ 2 ~ c?a; ~
8. (1 a; 2 ) j^ x ~j- = 0, where x = sin 2.
9. x* d ~ + 2* 2 |^ + 1; = 0, where a; = e".
d 2/ y 2 c?v v . .
10. 4. - - - _ 4. - - = o, where x tan z.
dx* T 1 z *
CHANGE OF VARIABLE 21
11. Transform the formula for radius of curvature,
N!)T
p= w %
dx*
into polar coordinates, the equations of transformation being
x = r cos 6, y = r sin 6.
1. ^
ANSWERS
d*x dx
dx
X --
df
z n z . o <? n z z ^ z ,
5. T-O 2 j- = sm 2 2.2 6. - 7 - - 2^-^ + 3^ 2 ,- +a; 3 = 0.
dx* dx * t dx* dx 2 n dx ^
A
- j-i j-2 + ^ = 0. 10. ,- + v = 0.
3 2 T a T
CHAPTER III
ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
AND FIRST DEGREE
20. An ordinary differential equation in one dependent vari
able, of the first order and first degree, may be represented by
the equation
Mdx + Ndy =
where M and N are functions of x and y and do not contain
derivatives.
The equation Mdx + Ndy = cannot be integrated in the
general form. There are certain particular forms of it, however,
which can be integrated. Some of these will now be investigated.
21. LINEAR DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
Definition. An ordinary linear differential equation of the
first order is an equation in the form
where P and Q are functions of x and do not contain y or deriv
atives.
The general solution of the equation
can be found as follows :
Multiply both sides of the equation by ef
22
EQUATIONS OF FIEST OEDER AND FIRST DEGREE 23
If the substitution u = yeJ pdx be made, the left hand member of
the equation reduces to du/dx.
..
dx
.-.y = e-f pdx fQef pdx dx + or/", (1)
which is the general solution of the equation.
In the original equation, if P is zero, the equation reduces to
the familiar form dy/dx = Q, and the general solution is
y = c + fQdx.
If Q is zero, the equation becomes
and the general solution is y = ce~J Pdx .
When Q, in the equation
is zero, the equation is called the ordinary linear differential
equation of the first order with the right hand member zero.
EXAMPLE. Find the general solution of the equation
dy 1
-T- + - y = x
dx^ x y
Multiply both sides of the equation by eJ * dx .
dy r l . dx 1
-
..~
dx
y =
9
24 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Let u = ye*/ * d *.
d u f-dx r 2 f-dxj
. - . -j- = #W * . . . u = I are- a# -f c.
efo J
J x 4 a? \ 4 / 4- *"
It is usual to solve an ordinary linear differential equation of
the first order by substituting directly in formula (1). Thus,
in the above example, formula (1) becomes
- fl d*
y =
22. EQUATIONS REDUCIBLE TO THE LINEAR FORM
A form easily reducible to the linear form is
where P and Q are functions of # and do not contain y or deriv
atives.
Divide by y n .
Let y w+1 = u.
du^ &*&&******
^ - -*.&.
C4 ^ "St**" 1
" c ft
ec^tferry*^!
,***-
which is linear and can therefore be solved by the methods of
Art. 21.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 25
EXAMPLE. Find the general solution of the equation
Divide by y*.
Let y~* = u.
. . -2y-* d ^=-
* dx dx
.-.- ?.--*-.
dx x
Therefore 2x*y* + ca; 2 2/ 2 = 1 is the general solution of the
equation.
23. VARIABLES SEPARABLE
Sometimes the equation Mdx -f Ndy = can be brought to
the form Xdx -f Fdy = where X is a function of x alone and
Y is a function of y alone. In such a case the general solution
is evidently
c being an arbitrary constant.
EXAMPLE. Find the general solution of the equation
x Vl y 2 dx -f- y Vl x*dy = 0.
Divide by Vl y z Vl # 2 .
3* ?/
= 0.
Vi-* 3 Vi-t/ 2
/ifc&
vrr?
Therefore Vl x* -f Vl 2/ 2 = c is the general solution of the
equation.
3
26 SHORT COURSE ON DIFFERENTIAL EQUATIONS
The process of reducing the equation Mdx -f Ndy = to the
form Xdx -f Ydy = is called separation of the variables.
24. EXACT DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
AND FIRST DEGREE
Definition. The ordinary differential equation Mdx + Ndy = Q
where M and N are functions of x and y, is said to be exact when
(there is a function u(x, y~) such that du = Mdx -+- Ndy.
EXAMPLE. The equation 2xydx + x z dy = is said to be
exact because u = x 2 y is such that du = 2xydx -f x*dy.
When there is a function u(x, y) such that du = Mdx + Ndy,
then u c, where c is an arbitrary constant, is the general solu
tion of the equation Mdx -f Ndy = 0.
Condition that the equation Mdx 4- Ndy be exact. If
the equation Mdx -f Ndy = be exact, then, by definition, there
is a function u(x, y) such that du = Mdx + Ndy. Now
7 du j du ,
du = -=- dx + -^- dy,
dx dy
from the definition of the differential of two independent variables.
,, du , , T du
.-.M=^-, and N= ^ .
dx y dy
dM d*u , dN d*u
and -T= = ^
dx dxdy
dM dN
dy ~~ dx
That the equation Mdx + Ndy = be exact, it is therefore
necessary that
dM dN
dy " dx 9
Conversely, the condition is sufficient. That is if
dM dN
dy == dx
then Mdx -f Ndy = is an exact differential equation.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 27
Proof: Let C Mdx = P, . . f- = M-
** ox
j^P dM _dN
dy dx~ dy ~~ dx
where Q(y) is such that dQ(y) = F(y}dy.
Therefore, if
_
dx - dy
the left hand member of the equation Mdx -f- Ndy = is an
exact differential and therefore the equation is an exact differential
equation.
To find the general solution of the equation Mdx -j- Ndy =
when the equation is exact.
Let u(z, y} be a function whose differential is Mdx + Ndy.
a- & u nr
Since = M,
dx
V-%r^*-
. . F(y} = \(N~
J \ dy
The general solution of the equation is u = c where c is
arbitrary constant.
^H
28 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXAMPLE. Find the general solution of the equation
O s -f 2xy 4- y}dx + (y* + a? + x)dy = 0.
This is an exact differential equation. Therefore the general
solution can be obtained by the above method.
~.
Since ^- = M,
dx
.u = f (af 4- 2xy 4- w)<&; + F(y) = T + tfy 4- ay -, _ w
/ 4 <tl
" W TT
ry =jr
K AV
. . dy
Therefore ~ -f ^ + ^y + \ c ls ^ e re( l u i re( l general solution
of the equation. .,
25. INTEGRATING FACTORS
It sometimes happens that the differential equation
Mdx + Ndy =
is not exact but becomes so when it is multiplied by some quan
tity. Thus,
of Art. 21, is not exact but becomes so after multiplication by
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 29
Definition. A factor which changes a differential equation
into an exact differential equation is called an integrating factor
of the equation.
Sometimes an integrating factor can be found by inspection.
EXAMPLE. Find the general solution of the equation
(<eV - y*)dx 4- 2xydy = 0.
The equation is not exact as it stands but becomes so on multi
plication by l/# 2 .
Multiply by 1/x*.
zV - f 7 2v ,
. * . - i-^- dx 4- dy = 0.
x 2 x
~
X
...*+*().*
V s
4- - = c. . . y* = xe* + ex.
Therefore y z = #e x -|- ca; is the general solution of the equa
tion.
Rules have been devised for finding integrating factors in
many cases where they cannot be found by inspection. For a
discussion of them, the student is referred to Boole s, Murray s,
or Johnson s Differential Equations.
26. EQUATIONS HOMOGENEOUS IN X AND y
Definition. If M and N of the equation Mdx -f Ndy =
are both of the same degree in x and y and are homogeneous, the
equation is said to be homogeneous.
To find the general solution of the equation Mdx 4- Ndy
when the equation is homogeneous.
dy M
30 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Divide both numerator and denominator of -^ by x raised to
the power indicated by the degree of M or N.
Then every term in M and N is constant or in the forn^ of a
coefficient multiplied by some power of - .
Then
dy_ f (y\
dx~ J \x)
Let y = vx.
dv .
x fo + v =fW-
Therefore
dx dv
x
an equation in which the variables are separated, and can there
fore usually be integrated without difficulty. /
t
EXAMPLE. Find the general solution of the equation
(x 2 + f)dx - xydy = 0.
equa
7 ,
dx xy
Let y = vx.
dv 1 4- v 2 dv I -{- v 2 1
. . v 4- x -y- = . . . x -j- = - - 1> = -.
dx v dx v v
dx
. . vdv = .
x
. . v* = 2 log ca?.
Therefore y z = 2x* log ex is the general solution of the equation.
dii a.x 4- b.y -4- c.
27. EQUATIONS OF THE FORM -/ = -i iy
V + b # + C
The general solution of an equation in the above form can be
found as follows :
Let x = x + X Q , and y = y + y , where a/ and y are new
variables, and X Q and y are constants.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 31
Change the variables to x and y r .
_
dx a/ + bjf
Case I. If X Q and ?/ can be determined such that
a i x 9 + %o + c i = > and % + tyo + C 2 = 0,
then, on determining them such, equation (1) becomes
dy a^ -f b,y
dx> ~ a,x r + &/
which is homogeneous and can be solved by the method of
Art. 26.
Case II. If x and y cannot be determined such that
a i x <> + b iy + c i = > and a ^o + b *y<> + c a =
then, as was seen in algebra,
5 _ *i _ 1
2 ~ 6 2 ~~ m*
By substitution, the original equation becomes
dx ~ m(a^x
Let aj -}- b^y = v.
. rfv dv
Therefore
C?V V -4- C.
j- = a, + o. - -,
J
an equation in which the variables are separable.
EXAMPLE 1 . Find the general solution of the equation
dy __ 6x - 2y - 7
dx = 2x + 3y 6*
32 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Let x = X Q + x f , y = y + y where
6z - 2y Q - 7 = 0, and 2z + 3y - 6 = 0.
_ 5_- 2v _ 6 - 4v _
= ~~ ~ ~~
3v
(2 + 3v)c?v dx
6 "TTSnTsl? = ^"
. . - log cX = log (3v 2 + 4v - 6)*.
Therefore 3y 2 -f- 4xy 6x* I2y -f- 14# = c is the general solu
tion of the equation.
EXAMPLE 2. Find the general solution of the equation
This comes under case II.
dy dv
. * . b 2 - = - .
.
-f 4 aa;
^ + 76
rfa; ~ v -f 8
+
. . v - 30 log (v + 38) = 2x + ,.
. . 4x - 2y -30 log (6* - 2y + 38) = c,.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 33
Therefore 2x y 15 log (3# y -f 19) c is the general so
lution of the equation.
EXERCISES
Find the general solution of each of the thirty-six following
equations.
dy x
*
dx jT = x(\ + z 2 )
6. <l_^)+(^-l)2/ = ^.
f/?y
7. -7- -f- cos a; y = J sin 2a;.
8. *(l-aog + (2*-l)f.*
9 s in ^/ = / 2 sina:. ^ 10. 1 - ^ 2 - / =
11. -j- +cosxy = 2/ n sin2a;. "12. 3?/ 2 -^ + y 3 = a; - 1.
13. - tan x y = y* sec #.
14. y sf~^ldx + a; ^f^ldy = 0. _
15. (e y + 1 ) cos # eta -|- ^ sin x dy 0.
16. V2cw/ ?/ 2 cosec xdx -\- y tan .T c?y = 0.
17- 2/(3 + 2/) = < 2 + 3). \
18. _
34 SHORT COURSE ON DIFFERENTIAL EQUATIONS
19. (V 4 4:xy 4- y z )dx 4 (2# 2 4- 2xy 4 ty*)dy 0.
20. sin x cos y dx 4 cos re sin y dy = 0.
,s
21. fz 2 .
22. a;(aj - 2y)dy + (> 2 + 2y*)da; = 0.
23. Sxydy - (a? + if)dx = 0.
24. O 2 + 3a^ - 2/ 2 )c?2/ - 3^daj = 0.
25. (x 2 4. 2iry)dy - (3^ 2 - 2sy 4. y*)da; = 0.
26. 5xydy - (x* + y 2 )^ = 0.
27. (ic 2 - 2xy)dy + (a; 8 - 3y + W)dx = 0.
28. &td 2x 2 _ 3*(te = 0.
29. (3z 4 2w - 7) / = 2z _ 3y 4 6
\ I */ / >y/>
30. (6a; - 5y 4 4) ^| = 2x - y 4 1.
U
31. r5^_2v47)^=a:-3v42.
^2/ fr
^\
dll n
33. (x - By 4- 4) /- = 2o; _ 6y + 7.
y - T y rfa;
/*
c?v
34. (5a; 2y 4- 7) g = lOo: -4^ + 6.
35. (2*_22,45)2- = *-</ + 3.
36. (6a? - 4^ 4- 1) = 3a?-
The following formulas, derived in almost any work in cal
culus, are inserted here for convenience of reference :
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 35
The subtangent and subnormal at a point (#, y) on a curve
whose equation is expressed in rectangular coordinates are
y T- and y -j- respectively. The polar subtangent and polar sub
normal at a point (r, 6} on a curve are r 2 -y- and -^ respectively.
The angle between the radius vector to a point (r, 0) and the
tangent line to the curve at the point is
ST
de
The equation of the tangent line to the curve y =/(#) at the
point (x v y^) on the curve is
dy
The area enclosed between the curve y =/(#), the a>axis, and
the ordinates whose abscissas are x and x l respectively is
Jyd*
provided the curve does not cut the #-axis between x and x r
The length of the arc of the curve y = f(x) between the points
(^o> 2/o) an( ^ (^iJ y\) on ^ ne curve is
37. Determine the curve whose subtangent at a point on it is
n-times the abscissa of the point. Find the particular curve that
goes through the point (3, 4). Plot the curve (a), for n = 1,
(6), for 7i = 2.
38. Determine the curve whose subtangent at a point on it
is ?i-times the subnormal at the point. Find the particular curve
that goes through the point ( V^, 2). Plot the curve when
36 SHORT COURSE ON DIFFERENTIAL EQUATIONS
39. Determine the curve whose subtangent is constant and
equal to a. Plot the curve, (a), when a = 1, (6), when a = 2.
40. Determine the curve whose subnormal is constant and
equal to a. Find the particular curve that goes through the
point (1, 2).
41. Determine the curve which is such that the length of the
perpendicular from the foot of the ordinate of any point on the
curve to the tangent line at that point is constant and equal to a.
Determine the particular curve when c = a. At what angle
does this curve cut the ^/-axis ?
42. Determine the curve which is such that the area between
the curve, the re-axis, and two ordinates, is equal to the arc
between the ordinates.
43. Determine the curve which is such that the perpendicular
from the origin upon any tangent line is equal to the abscissa of
the point of contact.
44. Determine the curve in which the angle between the radius
vector and the tangent line is ^i-times the vectorial angle. Plot
the curve when n = -J.
45. Determine the curve in which the polar subnormal is pro
portional to the sine of the vectorial angle.
46. Determine the curve in which the polar subtangent is pro
portional to the length of the radius vector.
The equation for a circuit containing induction and resistance is
di .
where e is the electromotive force [E.M.F.] impressed upon the
circuit, JR the resistance offered by the circuit, L the coefficient
of induction, i the current, and t the time during which the cir
cuit is in operation. In each of the four following exercises,
determine the current in the circuit after a time t supposing that
the resistance and induction are constant.
47. The E.M.F. is zero. Solve subject to the condition that
i = Jwhen t = 0.
48. The E.M.F. is constant and equal to E.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 37
49. The E.M.F. is a simple sine function of the time,
= E sin <ot where E is the maximum value of the impressed
E.M.F., and <o is the angular velocity, equivalent to 2-n-n where
n denotes the number of complete periods or alternations per
second.
50. The E.M.F. is the sum of two components each follow
ing the sine law, that is, e = E l sin ut -\- E 2 sin (but -f- 0).
The equation for a circuit containing resistance and capacity is
di i 1 de
where e is the E.M.F., E the resistance, C the capacity, i the
current, and t the time during which the circuit is in operation.
In each of the two following exercises determine the current
in the circuit after a time t, supposing that the resistance and
capacity are constant.
51. The E.M.F. is constant and equal to E.
52. The E.M.F. is a simple sine function of the time,
= E sin at.
The equation for a circuit containing resistance and capacity is
R dq q
ll di + ~c =
where e is the E.M.F., R the resistance, C the capacity, q the
quantity of charge in the conductor, and t the time during which
the circuit is in operation. In each of the three following exer
cises determine the charge in the circuit after a time t, suppos
ing that the resistance and capacity are constant.
53. The E.M.F. is zero. Solve subject to the condition that
q = Q when t = 0.
54. The E.M.F. is constant and equal to E.
55. The E.M.F. is a simple sine function of the time,
= E sin wt.
ANSWERS
/Y
1. 4xy = 2# 2 x* -f- c. 2. y sin x = log tan - -f. c.
38 SHORT COURSE ON DIFFERENTIAL EQUATIONS
3. y = x 1 -j- ce~ z . 4. y = x 1 i
5. 2/(l + a 2 )* = log -rp-
6. 2/= log (1 a 2 ) -fez. 7. y = sin a; 1 -j-
2i
8. y ax 4. cz Vl # 2 - 9. - = 1 -f ce~ cosa! .
"
10. i= _ a + c Vl - x\
y
11. y 1 " 71 = 2 sin a; - j-^ + ce~ (1 - n) 8ln z .
12. 7/ 3 = a; - 2 + ce~ x .
13. y~ s = 3 sin # cos 2 x sin 3 a; 4- c cos s x.
14. V# 2 f sec" 1 .r -f V^ 2 1 sec" 1 ?/ = c.
15. (e* -f 1) sin x = c.
_ nt
16. cosec# -|- A 2ay y* avers" 1 - = c.
17. 2y + Qy - 9 log (2y + 3) = 4^ 2 + c.
is. _ 5
19. *-r + 2rc 2 2/ + xy 1 + y 4 = c. 20. cos # cos y = c.
21. -^ -f a; log 2/ = c. 22. #V = c( -|- y) 3 .
23. (4^ - x>? = . 24. / = c (^~ )
25. etf + 3^ - 3, ) =
26. a - 1 = cz.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 39
[6w- (3 +
28. ex = \^
}6y - (3 -
29. (y - ) 2 + 3<> - AX? - if - s - = c -
30. (5y - 2* f 3) 4 = c(4y - 4a? - 3).
{2(y - A) - (4 -
_ _ c
32. (By - 5x + 10) 2 = c(y - a; + 1)-
33. 15y - 30^ + c = 3 log (5s - 15y + 17).
34. 4x - 2y + c = 16 log (5a? - 2y + 23).
35. 2# a; + c = log (a y + 2).
36. 2y - a; + c = J log (12a; - 8y + 1).
37. 2/ M = cz; 3/" = i w . 38. y=
Vn
39. y a = ce x .
40. 2 = 2cw; + c ; ?/ 2 = 2a# _|_ 4 2a.
41. y = {e 4- ? e j ; y - ^ C e " + e * ) ; zero
44. r n = c sin ?i0. 45. r = c k cos 0.
46. r = ce"*. 47. i = Je~^*.
48. i = | + c^ . ; .
E /E . \ -?-t
49. i = 7-^ r ( Y sm ^ to cos o>H 4- ce *
SHORT COURSE ON DIFFERENTIAL EOUATIONS
7?
50. i ~~~ - ~ s ^ n ^ w cos
]? \~ T> ~]
+ r^ - v tT-.ffli(fcrf+t) fciwiCWHrf)
i+W 1 -
52. i = ce ^ -j- . --- ^r^r-i (cos wi _|_ JtCw si
J. -j- _/x w
53. 5= Qe~^ . 54. 5=
55. g = , --- M ^ - - (sin tt RCu cos orf) 4-
1 -]- .it O (o
CHAPTER IV
ORDINARY LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS
28. Definition. An ordinary linear differential equation is
a differential equation in one dependent variable which is linear
in the dependent variable and its derivatives.
We saw in Art. 21 that the type of an ordinary linear differ
ential equation of the first order is
where P and Q are functions of x, and do not contain y or de
rivatives.
In genera], the type of an ordinary linear differential equa
tion is
d n _ d n ~ l _ d n ~*
where P v P 2 , , P n , and X, are functions of x, and do not
contain y or derivatives.
In this chapter the only cases considered are those where
P,, P 2 , , P n are constants and real. Two forms of this equa
tion present themselves, namely, when the right hand member is
zero, and when the right hand member is not zero.
RIGHT HAND MEMBER ZERO
29. We shall first prove a theorem used in the investigation
of equations in this form. It is :
Theorem. If y = y v y = y t , , y y n , are solutions of the
equation
d n y d n ~ l y
41
42 SHORT COURSE ON DIFFERENTIAL EQUATIONS
then
arbitrary constants, is also a solution of the equation.
Proof. Substitute y = c l y l -f c a i/ 3 -f 4. cjj n in the equa
tion.
Now each expression in brackets is zero, since
y = y v y = y y = y
are solutions of the equation. Therefore
is a solution of the equation.
Cor. If y = c l y l 4. c 2 t/ 2 4. -f c n y n is a solution of the equa
tion, then y = c^, y = c 2 y 2 , , y = e n y n , are solutions of the
equation.
If y = y lt y = y z , , y = y n are linearly independent solu
tions of the equation, then y = c 1 y l -f c a y 2 + -f c n y n is the
general solution of the equation (see Art. 13).
30. To find a solution of the equation
in the form y = e"" 5 .
Let y = e mx and substitute in the equation. If y = e is a
solution of the equation, then
+ P a m n - 2 + + PJ = 0.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 43
Since e mx cannot be zero for any value of m, then must
P = 0.
Therefore, if y = e mx is a solution of the equation, it is necessary
that
m n + Pjm"- 1 + P 2 m n ~ 2 + 4- P n = 0.
Conversely, if m has a value m x such that
7V + ^x*- 1 + J>r~ 2 + - + P* = o,
then t/ = e mia; is a solution of the equation. This is obvious be
cause on substitution of y = e mix , the equation reduces to
<pi*(mf + PjWij"- 1 + Pjn^ + - + PJ = 0.
Therefore the necessary and sufficient condition that equation (1)
has a solution in the form y = e* is that m be such that
Definition. The equation
m n + Pjm"" 1 + P 2 w n ~~ 2 H ---- + P n =
is called the auxiliary equation of
31. To find the general solution of the equation
n n ~ l l ~*
When the auxiliary equation has distinct roots. Denote the
roots by m v ?7i 2 , , ra n . Then n linearly independent solu
tions of the equation are y = e mix , y e m * x , , y = e m " x , and
the general solution is y = c^ 1 * 4- c^e** -f -f c n e mnX (see
Art 29).
44 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXAMPLE 1. Find the general solution of the equation
g-3^-4y = 0.
dx* dx
Let y = e.
2 _ 3m - 4) = 0. . . (m 4)(m + 1) = 0.
Therefore y = c v e ix -f c t e~* is the general solution of the equation.
EXAMPLE 2. Find the general solution of the equation
Let y = e mx .
. . ^(m 2 + m + 4) = 0.
1 -
^
Therefore y = ^e a * + c^e a * is the general solution of
the equation.
This solution may be written as
Vl5
y = e^""*" cos ^ a; -|- c a e~ 4z sm ^- # (see Art. 5).
When the auxiliary equation has multiple roots. Suppose
that the auxiliary equation m n -f-P 1 m rl ~ 1 -|-P 2 m n ~ 2 -j- +P n =
has the roots m v w 2 , m y , m n .
At first suppose that two roots are equal. Suppose for defi-
niteness that w 2 = m,. Then a solution of the differential equa
tion is
y = (c, + c,)^* + c 3 e-3x + Gne ^.
Since c x + c 2 is equivalent to only one constant, this solution con
tains only Ti 1 arbitrary constants and is not therefore the gen
eral solution of the equation.
To find the general solution in this case :
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 45
Suppose that the differential equation is such that its auxiliary
equation has the roots m v m l 4- h, m y , m n . The general
solution of this equation is
y = c^ x 4- c+* 4- c z e m ** 4. ----
= c^* 4- 0,6^*6** 4- c s e m ^ _j_ ---- ^_
Expand e** by Maclaurin s Theorem to TI terms and the re
mainder.
< x l < x.
Substitute in the above equation.
Since c 2 is arbitrary, ^ may be chosen such that cji is any con
stant B for all values of h. Since c, is arbitrary, c x 4. c 2 may be
chosen such that c x 4. c 2 = A. Then
4-
where J. and i> are arbitrary constants.
Let h approach zero. As h approaches zero, the assumed
auxiliary and differential equations approach identity with the
given ones, and (1) approaches the general solution of the given
differential equation.
Now
Therefore the general solution of the differential equation is
y = (A 4- Bx^e*** 4- c s e TO ^ _|_ . . . _j_ c n e m x ,
46 SHORT COURSE ON DIFFERENTIAL EQUATIONS
or, as we shall write it,
y = (i + W)e m * 4- c s e m & H + c n e m * x .
In a similar manner it can be shown that if three roots of the
auxiliary equation are equal, the general solution of the differen
tial equation is
and, in general, if r roots are equal, the general solution of the
equation is *
y =. (c, 4- c,x 4- 4- c x r ~ l ^e mix 4- c , e mr+lX 4. j_ c e mnX
J \ 1 l 2 t r / ~ r+1 I n
If a pair of imaginary roots occur twice, the part of the general
solution derived from these roots is
s px+j sin fix) 4- ( c 3 -\-c 4 x)e aX (cos (3xj sin fix)
Lj 4- -#!#) cos /3^ 4- ( A 3 4- -B 2 a:) sin fix] .
EXAMPLE 1. Find the general solution of the equation
c? 2 v ^ dy
y _l_ 2 -^ 4. i/ = 0.
The auxiliary equation is m 2 4- 2m 4- 1 = 0, or (m 4- I) 2 = 0.
The general solution is therefore y e~ x (c l 4- c 2 #).
EXAMPLE 2. Find the general solution of the equation
The auxiliary equation is m* 4m 3 4- 8m 2 8m 4-4 = 0, or
The general solution is therefore
y = e x {(A l 4- Bfl) cos x 4- (J. 2 4- J5 2 #) sin x}.
32. As a physical application of the above principles, con
sider the following discussion (see Emptage, Electricity and
Magnetism, page 180) :
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 47
In a galvanometer in which resistance is offered to the motion
of the needle, the equation of motion of the needle for small
oscillations may be written as
g + 2*| + ^-.)=0, (1)
where is the angle through which the needle turns in the time
t, k is a constant depending on the resistance offered to the mo
tion of the needle, w 2 is a constant depending on the moments of
the restoring forces on the needle, and a ws the angle which the
needle at rest makes with the line from which angles are meas
ured. Let 6 a = , and substitute in (1).
d*6 )7 dff
..y + ii-j+rfr.a
This is a linear differential equation of the second order with
constant coefficients and right hand member zero. The auxiliary
equation is m 2 -f- 2km -f to 2 = 0. The roots of the auxiliary
equation are m = k
Case I. If k > o>.
In this case, - a = e^-*^** 1 ^) + C j>(-k-vi^w is tlie
general solution of (1).
Case II. If k = w .
In this case a = (c x -f- c.)e~ ltt is the general solution of
(1).
Case III. If k < <o.
In this case a = e~ w [c 1 cos V^ 2 tf t + c 2 sin V^ 2 k 2 1]
is the general solution of (1).
In cases I and II the motion is not oscillatory. The needle
can go through the position of equilibrium for one value of t,
after which it reaches a position of maximum deflection and then
continually approaches but never reaches the position of equi
librium. In case III there are oscillations in equal times, the
periodic time being
48 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXERCISES
Find the general solution of each of the following equations.
* .a e.
V ANSWERS
1. y = c^- 2 * + cf. 2. y =
3. y = c^ + c,e 2a: + c s . 4. y = e~ x (e l
5. y = e z (X + c 2 a; 4. c 3 z 2 ).
6. y = ^e* -f- e~ 2a! (c 1 cos ^ -j- c 3 sin ).
7. y = CI P "*" + <5 z (c 2 cos x -\- c, sin a;).
8. y = Cje* 4- c.,!?" 35 4. c 3 cos a; 4 c 4 sin x.
9. y = (Cj 4 CjZ) cos a; 4- (c 8 4. c 4 x~) sin a;.
RIGHT-HAND MEMBER NOT ZERO
33. Symbolic form of equation. The equation, when the right
hand member is not zero, is
where P,, P 2 , P n , are constants, and Xis a function of x but
not of y.
Let
d D
dx -"
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 49
and the equation may be written,
D n y -f PJ) n ~^y -f P 2 D n ~ 2 2/ -f + P n y = X.
Suppose that y is treated as an algebraic factor of the left hand
member of the equation. On this supposition, the equation
becomes
Suppose also that V" + P l D H ~ l + P 2 Z> W ~ 2 + - . . + P n , factored
as an algebraic expression in D, is
and that the equation is written
(D - m^(D - mj . . - (D _ m^y = X (2)
Equation (2) is not equivalent to equation (1) except in a
symbolic sense. Let us see what conventions must be made in
order that equation (2) be equivalent to equation (1).
Let us make the convention that (D m)u where u is a func
tion of x is equal to
du
- _ mu.
dx
Also, let us agree that we shall begin at the right of the left
hand member of (2) and work towards the left, evaluating
according to the preceding convention at each step. Then
(D - m^XD - mjy = (D _ m^) _ m.
50 SHORT COURSE ON DIFFERENTIAL EQUATIONS
and finally,
(D_m 1
d n y
Now -(X + m 2 + ... + wi B )=P 1 , ., (
since the factors of D n + P^" 1 -f P 3 D n ~* 4. - + P n , treated
as an algebraic expression in D are D m v Dm v , Dm n .
This expression is therefore the same as the left hand member of
equation (1). Therefore, with these conventions, equations (2)
is equivalent to equation (1).
EXAMPLE. With the above conventions, the equation
may be written in the equivalent form (D m 1 )(Z)_m 2 )y = X,
where D m, and D m 2 are the factors of the expression
D 2 4- PJ} -f- P 2 treated as an algebraic expression in D. For,
^^ ^V
(D - mjy = ^ - m 3 7/,
(D - m^D - mjy = (D - m
Now (Wj -|- m 2 ) = P p and m^m^ = P r Therefore the second
form of the equation is equivalent to the first.
Definitions. When equations in the form (1) are expressed
in the form (2), they are said to be expressed symbolically, or
to be expressed by means of symbolic factors.
When a symbolic factor D m and a function u are applied
to each other so as to give (D m)w or -=- mu, the function
do&
u is said to be operated upon by D m, or the factor D m to
be multiplied symbolically by u.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 51
The factor D m is called the symbolic operator, or more
briefly, the operator.
34. Theorem. The order in which the symbolic factors in the
equation of the last article are taken is immaterial.
Consider in illustration the equation of the second order.
Let the equation be taken in the form
(D - ro f ) (D - mjy = X.
Then
(D m ") dy - m
and
(D-m,)(D-m 1 )^=(D_m ]
= *2 _ (m 4- m ^ 4- %
dx 2 ^ i ~f *J dx "*"
Therefore (D m 2 )(Z> m^)y = X is equivalent to
Also, (J> m 1 )(D m. 2 )y = X is equivalent to
S+ P + P ^ = X - (See Art. 33.)
Therefore, in the case of the equation of the second order, the
order in which the factors are taken is immaterial.
The proof in the general case is left as an exercise to the
student.
35. First method of solution of the equation
(D-mJCD-m,)? = X.
Let (D m^y = u. The equation then becomes
(D m l )u = X or - _ m^u = X.
52 SHORT COURSE ON DIFFERENTIAL EQUATIONS
du ^
_ m.u = JC
dx l
The general solution of the equation
is (see Art. 21)
. .(D-
u =
This is the general solution of the given equation.
EXAMPLE. Find the general solution of the equation
d 2 y n dy _
-=-5 3 -/ + 2y = cos a;.
ciic 2 rfa;
Write the equation as (D 1)(Z> 2)y = cos a?.
Let (D 2)y = w. The equation then becomes
..p.
( l> 1 )u = cos a? or -=- u = cos a;.
c?a;
. . u e* J* e~* cos x dx + c^*
J(sin x cos x) -|- Cje*.
. . (Z> 2)y = J(sin a; cos a;) -f ^e*.
. . y = Je 1 * J*e~ 2a! (sm a; - cos x)dx + c^* f e~ x dx + c a e te
= T V cos a: T \ sin a; + c^ + c/ 2 .
This is the general solution of the given equation.
36. To solve the equation
(D - m,}(D - m,) - (D _ mjy = X,
we may proceed as follows :
First, let ( D m 2 ) - (D mjy = u. The equation then
becomes (D m^u = X.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 53
From this equation, u can be determined as in the case of the
equation of the second order. Let
Then
(D _ m^v = u.
From this equation v can be determined in the same manner as
u was determined before. After n 1 such steps there results
(Z) m n )y = 2 where 2 is a known function of x.
The general solution of the equation (Z> m n )?/ = 2 is the
general solution of the original equation.
37. The following theorems concerning the symbolic operator
will now be established :
Theorem I. A constant factor in a function may be written
in front of the operator.
Proof: Let au be a function containing a constant a as a fac
tor. Let D m be the operator. Then
(D m)au = = mau, by definition
du
Theorem II. The result when the operator is applied to the
sum of a number of functions is equal to the sum of the results
found when the operator is applied to each of the functions
separately.
Proof : Let u -\- v + w -}- -{-2 be the sum of a number of
functions. Let D m be the operator. Then
(D _ m) (u -f v -f w -f -f z)
d(u -f v -f w -f-
dx
m(u -f v + w -f -j- 2)
du dv dw d
mv-f-y- mw -f -f -j
r dx ~ d
m)v -f (D w)w -f + (Z) tri)z.
--y- --y- mw -f -f -j- _ mz
dx dx r dx ~ dx
54 SHORT COURSE ON DIFFERENTIAL EQUATIONS
38. The equation (D - mj (D - m a ) (D - mjy = X
may be written in the form
- ro.) (D _ mj
In the first form the symbolic operators
D m 1? D m 2 , , D m n
applied in succession give X. Moreover, by the theorem of
Art. 34, the order in which the operators are applied is imma
terial. If the second form, therefore, is to be the same as the
first, the symbolic expression 7-= -- ^7^ -- -r -- 7-^ -- r X
(D-mJCD-mJ (D-mJ
must be such that, when operated upon by
D m v D m 2 , , D m n ,
in succession in any order, the result is X.
Definition. The symbol 7-= - r-^ r -- -7^ - r- is
(D _ m^(D _ m 2 ) - - - (D - mj
called the inverse symbolic operator, or, more briefly, the in
verse operator.
39. Let
be a linear differential equation where the symbolic factors viewed
as algebraic factors are distinct. Break up -^
r
into partial fractions as if it were an algebraic expression in D.
Then
(D mJCD m f ) ^ wij - m a \ I) - m l ~ D - m 2
Let
- ^ - X = w and - m ^_ m D ^_ m X=v>
Theorem. The result of operating on u + v with
m.) is X.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 55
Proof: Operate on u -j- v with (D m^^D m t ).
- m> +
(D m 1 )(Z> ra 2 >, by theorem II, Art. 37.
Now
(D-m 1 ) = -^--X ) and (D - m> = _ _J l - X,
lit/ ^o tit, "t/rt
by definition and theorem I, Art. 37.
= (D _ m 2 ) ^X+ (D - m.) f .
N A S A^J *,* IX lX\ /v - .
40. When the symbolic factors D m l and D m 2 , viewed
as algebraic factors, are distinct, the result of operating on
with (D m l )(D w 2 ) is JT, by the preceding article, and the
result of operating on
x
with the same factors is X, by definition. Therefore when the
symbolic factors D m l and D m 2 , viewed as algebraic fac
tors, are distinct, the inverse operator of
may be broken up into partial fractions the same as if it were an
algebraic expression in 7), and the result of operating with
(D _ m^^D w 2 ) on the expression formed by multiplying
each of the fractions symbolically by X, and taking the algebraic
sum of the results, is X.
56 SHORT COURSE ON DIFFERENTIAL EQUATIONS
In general, when the symbolic factors
viewed as algebraic factors are distinct, the inverse operator of
1 X
y = (D - mJCD - ro f ) (1> - m.)
can be broken up into partial fractions the same as if it were an
algebraic expression in Z), and the result of operating with
(D m l )(D - m a ) (D - m n ), on the expression formed by
multiplying each fraction symbolically by X and taking the
algebraic sum of the results is X.
The proof of this theorem is left as an exercise to the student.
41. Second method of solution of the equation
Break up __ r into partial fractions the same as
r (D -mJ(D m 2 )
if it were an algebraic expression in D.
. i JL_ _/__!_ J_Y
(D _ wJGD - m 2 ) ~ m t - m a \D - m t D-mJ
Let
w -- -=: - X and v = - w X.
~ mi m 2 /) m l m l m 3 JJ ^ m a
Operate on u with D m r
du
. . -r-
dx
Operate on v with D m a
. . v = _ _ ^ - e** C
m x m 2 ^
e~ m * x Xdx 4. c,d".
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 57
^
__ - _ e i* CtrwXdx - - - - &* Ce~ m - x Xdx
y m l m i J m l m 2 J
+ Cje"!* + 0,6""*,
ivhich is the general solution of the equation.
EXAMPLE. Find the general solution of the equation
d?y r.dy
^ _ 3 -4- 2y = cos #.
dz 3 do; T
Write the expression in the form
Break up =r - YTn - 9^ * nto P art ^ a ^ fractions the same as if
it were an algebraic expression in D.
2) - ~ ^ITT D-2
Let
-F^ T
jt> 1
cos x = u an " cos x = v>
Operate on u with D 1.
eft*
. . -,- u = cos x.
dx
. - . u = ^ cos x \ sin x -f- c^e*.
Operate on v with D 2.
ey
. . _ 2v = cos re.
dx
. . v = | cos x 4- | sin a; -f- c 2 e 2z .
. . y = T V cos a? T \ sin a; + c^ 36 -f
which is the general solution of the equation.
This method does not apply when the symbolic factors viewed
as algebraic factors are not distinct.
5
58 SHORT COURSE ON DIFFERENTIAL EQUATIONS
42. It will be noticed in the example of the preceding article
that the result is the same as that found by applying the method
of Art. 35 to the same equation. This will be the case in any
linear differential equation with constant coefficients to which
both methods apply.
The first method of solution will apply in all cases where the
left hand member of the equation can be factored into linear fac
tors in D. The second method will also apply if the linear fac
tors in D are all distinct. If two or more factors are equal, and
the inverse operator be broken up into partial fractions, the term
or terms corresponding to these factors may be evaluated by the
first method.
Usually the second method is easier of application than the
first.
43. An examination of either method by which the general
solution of a linear differential equation of the nth order with
constant coefficients and second member not zero is derived shows
immediately that the general solution consists of the sum of two
parts, one containing terms not involving arbitrary constants,
the other containing terms involving such constants. Moreover
the arbitrary constants are involved so that when any one is zero,
the term in which it appears vanishes.
Definition. The part of the general solution of a linear dif
ferential equation with constant coefficients and second member
not zero which contains the arbitrary constants is called the com
plementary function of the general solution of the equation.
EXERCISES
Find the general solution of each of the fourteen following
equations.
LINEAE EQUATIONS WITH CONSTANT COEFFICIENTS 59
d 3 y _ d 2 y _ dy
dx
.
9. ^, + ^
^ 2 2/
=
12. j4 j^ -f T^ V= cos a.
3 2 ^ *
In each of the six following exercises, find the equation of the
elastic curve of the beam from the given differential equation,
determining the constants of integration. Find also the deflec
tion of the beam. In these equations, E is the modulus of elas
ticity, I is the moment of inertia of a cross section of the beam
about a gravity axis in the section perpendicular to the applied
forces, and I is the length of the beam.
15. The beam rests on supports at its ends.
weightless with a weight P at its middle point.
P/l
It is supposed
60 SHORT COURSE ON DIFFERENTIAL EQUATIONS
16. The beam rests on supports at its ends. It is supposed to
be of uniform cross section and of weight w per unit of length.
Y
17. The beam rests on supports at its ends. It is supposed to
be of uniform cross section and of weight w per unit of length,
and to have a weight P at its middle point.
P
X
18. The beam is a cantilever fixed horizontally in the wall,
It is supposed weightless with a weight P at its extremity.
19. The beam is a cantilever fixed horizontally in the wall.
It is supposed to be of uniform cross section and of weight w per
unit of length, and to have a weight P at its extremity.
Y
dot
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 61
20. The beam is vertical. It has rounded
ends. It is supposed weightless. It is de
flected a small amount a and a load P is
applied at its upper end just sufficient to hold
it in position.
The equation for a circuit containing re
sistance, induction and capacity in terms ol
the current i is
L dt
in terms of the quantity of charge q is
3
*g
dt*
Edq
~Ldt
where e denotes the E.M.F., R the resistance, L the induction,
C the capacity, and t the time during which the circuit is in
operation. In each of the three following exercises, determine
the current and quantity of charge in the circuit after a time t,
supposing that the resistance, induction and capacity are constant.
21. The E.M.F. is equal to/0). Solve when R*C? 4L.
22. The E.M.F. is constant and equal to E.
23. The E.M.F. is a simple sine function of the time,
= .Esin tat. Solve when R*C ^ 4L.
ANSWERS
. y = T Z
2. y = ^ -
3. y = ^t 2
(3 T/13 )a!
62 SHORT COURSE ON DIFFERENTIAL EQUATIONS
5. y = T V* 2 - $p + Cj + c t e~ zx + c 3 e- 8a6 .
6. /=-
i
7. y = e x (|z 2 + c i + <yc). 8. y = a? + 2 + 6*0,
n a; sin a?
9. y = <r 4- Cj cos a; + c 2 sin a?.
10. y = - \x + J 4- Cl e- z + c 2 e~ 2j: + cf.
x cos a;
11. y = ^ -f c x cos a; -j- c 2 sin a;.
12. ?/ = J(cos x x sin x x cos #) -f- Cj sin a; -f- c 2 cos a; -j- c 3 e z .
13. y = e x (%x* + eX + c a a; + c 3 ).
14. y = .^ 4 24-|- Cj6 z -f- Cje" 2 -f c 3 sin re 4. c 4 cos x.
15. 4E = ^ - ^ 8 . Deflection = .
16. SJSTy = " ** - f *<. Deflection = .
wl + P , w (wl
17. -
IP?
3
18. 2^Jy = _ Pfo 1 + a: 3 . Deflection = r .
Deflection = pl3 + wl> -
20. ^
a
Deflecti
\~P I D ^/T 2
ion = a vers \\^rni anc * ** = ~72~
\ xi/7 .2 t
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 63
21 -
where
2LC
The value for q differs from that for i only in having /() instead
of/ (0- _
RC-Jl&C*-4LC
22. i = c"" * LG ce
cos
sin ^ ~2LC- -<>
when J?C=4JD.
when
cos
v 2LG
when R*C<4L.
q=
when R*C=4L.
23. i=
64 SHORT COURSE ON DIFFERENTIAL EQUATIONS
where T^ and T 2 have the values given in exercise 21.
CHAPTER V
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS. EXACT
LINEAR DIFFERENTIAL EQUATIONS
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS
44. Definition. A homogeneous linear differential equation is
an equation of the form
d n y n . d n ~ l y dy ~
f - x = x
where p v p v , p n _ v p n are constants, and X is a function of
x but not of y.
This equation can be transformed into an ordinary linear dif
ferential equation with constant coefficients by changing the inde
pendent variable from x to z, the equation of transformation be
ing x = e*. The equation that results from the transformation
may be solved by the methods of the last chapter. If a solu
tion is y =/(), the corresponding solution of the original equa
tion is y =/(logz).
45. The transformation and general solution of a homogene
ous linear differential equation in the general case will not be
considered here. We shall merely consider them in a particular
example.
EXAMPLE. Find the general solution of the equation
x-7/^s i c\ y^ (\ y
Let x = e?. . . z = log x,
dy dy dz 1 dy
dx~ dz dx~ x dz
d*y d ( 1 dy \ 1 dy 1 d*y dz 1 ( d*y dy
65
66 SHORT COURSE ON DIFFERENTIAL EQUATIONS
^x 2 \dz 2 dz / I ~ x 3 \ dz 3 dz 2 dz /
dy dy
X dx = dz
d 2 y _ ffy dy
dx 2 ~ dz 2 ~ dz
x s A = A _ 3 <?y , 2 ^
Substitute in the equation.
d*y d*y dy
dz* "" dz 2 ~ dz" y ~
The general solution of this equation can be found by the
methods of the last chapter. It is
y = t z + + c i e + c a e + C 3 e
The general solution of the original equation is therefore
c
v = 1 log #-1-1-4- 4-~4- c a a; 2 .
J *k. O I 4: I . T^ ~2 I 3
EXACT LINEAR DIFFERENTIAL EQUATIONS
46. Definition. A linear differential equation
is said to be exact, when, if the left hand member be represented
by V, the expression Vdx is the exact differential of some func
tion U which does not contain an integral of y.
The expression U is evidently an expression actually contain
ing a derivative of order n 1.
47. To find the necessary and sufficient condition that the
equation of the preceding article be exact, and a method of solu
tion of such an equation.
EXACT LINEAR DIFFERENTIAL EQUATIONS 67
Multiply each term by dx and take the integral of each term.
Cp d " y dx Cp^dx CP d y dz
J r d*" d * + J p d^ d +J p - &*
Now
jP n ydy = JP n ydy identically.
And, by integration by parts,
J*P_, g dx = - fP ^ydx + P^,
where the primes denote differentiation with respect to x.
. . fxdx + c = /(P. _ P n _, + P" n _ 2 - P"U + -}ydx.
(P n _ ! -P n _ 3+
Write the expression in brackets as Q a , _,, , Q respec
tively.
e =Q n ydx+Q n _,y+ ./ + . . + ft. (1)
Now in order that the equation be integrable there must be no
term in the right hand member of (1) containing an integral of
y. The necessary and sufficient condition for this is that Q n = 0.
68 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Therefore the necessary and sufficient condition that the equation
be exact is that Q n = or
P. - -f".-. + ?"~. - P" *-* + = 0. (2)
When this condition is satisfied the equation reduces to
c. (3)
, If the coefficients in (3) satisfy a relation in Q similar to (2)
in P, equation (3) is exact and the above process may be re
peated.
EXAMPLE. Solve the equation
P9 p/ O p" in P " _ fi
ft = A * B _! = A -r n _ 2 = - lu, r n _ 3 = o.
P P j^ P r/ P " _
* J n-l ~T * n-2 * n-3 ^
The necessary and sufficient condition that the equation be exact
is therefore satisfied.
The equation therefore reduces to
In this equation, P n = 2x, - P n _! = 4, P" n _ 3 = - 6z.
P P -L_ P"
* ^n * n-l ~T * n-2 U
The necessary and sufficient condition that this equation be exact
is therefore satisfied.
_, = - 2** + 3x* - 1 = _ 1, ., = x(l _ ).
The equation therefore reduces to
EXACT LINEAR DIFFERENTIAL EQUATIONS 69
This equation is not exact. It is however an ordinary linear dif
ferential equation of the first order, and can therefore be solved
by the method of Art. 21. The general solution is
which is therefore the general solution of the original equation.
EXERCISES
Find the general solution of each of the following equations.
~ cot * + cosec2 ^ = cos x -
70 SHOET COURSE ON DIFFERENTIAL EQUATIONS
fJ^/ll (Xl]
12. O + 8* 1 ) ^ + 2(1 + te) J + 6y = sin a.
13. (* 3 + *_ 3* + 1)+ (9* + 6z - 9)
,
14. x* --
ANSWERS
|f log *
A
2. y = _ + c^ 2 + ^-.
3. y \y? log x %x* + c^ + c 2 o; 2 4. c 3 a;
4. y = Jlogz + i + c x a? + c 2 o; 2 + ^.
5. y = | + ~ 2 log* + c 3 a;. 6. y =
7. y = log x + 2 + Cjic log a; + c 2 a;.
a; 3 c.a; c
10. y == xsm x -\- ^ sin # log (cosec # cot ^) -j- c 2 sin
11. =
12. (a; + 3a; 2 )2/ = sin x 4- c t a; 4- c a .
13. (a; 3 4. z 2 - 3^ 4- 1 )y = yh^ 4- i + C 2^ +
CHAPTER VI
CERTAIN PARTICULAR FORMS OF EQUATIONS
48. An equation in the form ^ = f(x}
dx n J ^ J
An equation in this form is exact and can therefore be inte
grated by the methods of the preceding chapter. It can also be
integrated by direct integration.
The first integration gives
where a t is an arbitrary constant.
The second gives
/ /
= J J -
where a 3 is an arbitrary constant.
After n integrations there results
where c v c a , , c n are arbitrary constants.
49. An equation in the form J f(y )
dx n ~ J ^ J
An equation in this form can in general be integrated only
when n = 1 and n 2.
When n = 1 the equation is
71
72 SHOET COURSE ON DIFFERENTIAL EQUATIONS
To integrate, separate the variables.
When n = 2 the equation is in the form
3 =/<*>
To integrate, multiply by 2 .
Now
dx dx* ~ dx \ dx I
Suppose that
dy
. .
V^(y) + Ci
dy
50. An equation that does not contain x directly.
Such an equation is of the form
CERTAIN PARTICULAR FORMS OF EQUATIONS 73
Let
Then
d?y dp dpdy dp
dx* ~~~ dx ~ dy dx ~ P dy
d*y d I d 2 y \ d / dp\ d / dp \ dy
da? ~~ dx \ dx* / ~ dx\ P dy ) ~ dy \ P dy ) dx
and so on.
The equation then becomes a differential equation in p and y
of order n 1. Suppose that it can be solved and that the solu
tion is p =/(y). Then a solution of the original equation is
51. An equation which does not contain y directly.
Such an equation is of the form
Let
dy
The equation then becomes a differential equation in p and x
of order n 1. If the equation can be solved for p and the
solution is^> =/(#), a solution of the original equation is
y + c =
52. An equation of the first order solvable for y.
In such a case, when solved for y, the equation becomes
y = F(x,p}. (1)
74 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Differentiate with respect to x.
This equation does not contain y explicitly. It is an equation of
the first order in p and x. If it can be integrated as an equation
in p and x, there results on integration an equation between x, p
and an arbitrary constant. From the resulting equation and
(1), if p can be eliminated, there results an equation between
x, y and an arbitrary constant, which will be the general solution
of the equation.
53. An equation of the first order solvable for x.
In such a case, when solved for x, the equation becomes
y = F(y, p). Differentiate with respect to y.
The method of procedure from this point is similar to that in the
preceding article.
EXERCISES
Find the general solution of each of the twelve following
equations.
!.g = , .g-COB,
-3
ffy /dy\*
7 ql ?. I -2- I 1
y da? \dx) ~
CERTAIN PARTICULAR FORMS OF EQUATIONS 75
= 8*. 10. *=
dx dx
-L J. * SO ~~ ty T- I ~^j I -1. ^J . U 7~"^ *V i
1 -*- dx
13. Find the curve whose curvature is constant and equal
to/c.
14. If a sphere of radius E l is surrounded by a concentric
shell of radii R 2 and jR 3 , the potential function, V, at a point
either in the space between the conductors or outside the outer,
satisfies the equation
J2 rr o J ~rr
= o,
dr* r dr
where r is the distance of the point from the center of the sphere.
Solve the equation given that F x is the potential on the sphere
and F 2 on the spherical shell.
15. If a circular cylinder of radius R l is surrounded by a cir
cular cylindrical shell of radii Jf? 2 and J? 3 , both of very great
length, the potential function, F, in the space between the con
ductors, is such that
dr* ^ r dr ~
where r is the distance from the point to the axis of the cylinder.
Solve the equation given that V l is the potential on the cylin
der and F 2 on the spherical shell.
ANSWERS
2. y = cos x + CjX + c 2 .
3. y = - log x - |(log x} z 4. CJR 4- c a .
4. y = Cj sin (aa; 4- c a ).
76 SHORT COURSE ON DIFFERENTIAL EQUATIONS
5. d= x + c 2 = Ve# - 2 + .log
c i c i
6. y = c 2 e c i*.
7. a; + c x = -log (cy + VcV - 1).
- 2).
1 C, 2 4- 1 . ..
8. y = - ~x - -^-n log (1 -
9. y =
10. y + c x = %x* -
11. \x + c x ^ =p Vz - y - log (1 =F V^ - y).
12. y = ex c 2 .
13. A circle of radius -.
K
JZ,^ F.-7 J.F.
= ~
15. F=
CHAPTER VII
ORDINARY DIFFERENTIAL EQUATIONS IN TWO DEPENDENT
VARIABLES
54. So far, the differential equations considered consisted of
two variables, one independent and one dependent. We shall
now consider equations in three variables. These may be divided
into two classes : those in which there is only one independent
variable, and those in which there is only one dependent vari
able. The first comes under the class called ordinary or total
differential equations : the second, partial differential equations.
This chapter is taken up with a discussion of a few forms of ordi
nary differential equations. The next chapter is devoted to
partial differential equations.
55. If /(#, 2/) is a single valued and continuous function of
the two independent variables x and y, given by the equation
z == /(#, y), and and ^-are continuous, then, by definition,
ox oy
, dz , dz ,
dz = dx + dy,
dx dy "
or
If f(x, y, z) is a single valued and continuous function of the
three independent variables x, y and z, given by the equation
u = f(x, y, z), and , and are continuous, then, by defi-
ox uy uz
nition,
y,
d , &
dx dy dz
77
78 SHORT COURSE ON DIFFERENTIAL EQUATIONS
56. Equation (1) of the preceding article has, as a special
case when 2 = 0, the equation
That is, the equation is true for the equation f(x, ?/) = where
x and y are independent variables. If y in f(x, y) = is a
single valued and continuous function of x, the equation holds
true for all values of x for which y is a single valued and con
tinuous function, for in this case y is merely restricted to values
which it could assume, as well as others, in the more general
case where it is independent.
This can be seen more clearly perhaps by a consideration of
the geometrical representations of the equations.
The equation z = f(x, y) when x and y are independent vari
ables represents a surface. If z = 0, the surface is the zt/-plane,
and the equation
dx dy
holds true for every point in the plane. If y is a single valued
and continuous function of x, the equation f(x,y)=Q repre
sents a curve in the zy-plane in which the equation expressed in
the form y = <j>(x) gives a single valued and continuous function
of x, and since
dx dy
holds true for all sets of values of x and y in the plane, it holds
true for all sets of values which together determine a point on
the curve in the plane.
57. Equation (2) of Art. 55 has as a special case when z = 0,
the equation
K*JL1> dx + d J^>^ dy + d J^^ dz = o.
dx dy dz
EQUATIONS IN TWO DEPENDENT VARIABLES 79
By reasoning similar to that employed in the preceding article
in the case of two dependent variables, it may be seen that
this equation holds true when z is a single valued and continuous
function of x and y.
58. An integral relation in x, y and z, equated to an arbitrary
constant c, say <(#, y, 2) = c, can always be expressed in the
form
Pdx 4- Qdy + Rdz = 0,
where P, Q and R are functions of x, y and z, and do not con
tain the arbitrary constant c.
For, the result of taking the differential of each member of
the equation <(#, y, 2) = c is, by the preceding article,
dt(x y, z) S<j,(x y, z)
dx dy dz
and this equation is in the specified form.
EXAMPLE. The result of taking the differential of each mem
ber of the equation x*y xz 2 4- y z z = c where c is arbitrary, is
(2xy - z^dx 4- (3? 4- 2yz)dy + (y 2 - 2xz)dz = 0.
This equation is in the form Pdx 4. Qdy + Rdz = 0.
The resulting equation Pdx -f Qdy -\- Rdz = is such that P,
Q and R are proportional to
d <*> d <l> and <W
dx dy dz>
respectively.
Conversely, however, an equation of the form
Pdx 4- Qdy + Rdz =
where P, Q and R are functions of x, y and z, does not neces
sarily give rise to a solution of the form <f>(x, y, 2) = c. This
can be seen immediately because an equation of the form
Pdx -f Qdy 4. .Kdz =
80 SHORT COURSE ON DIFFERENTIAL EQUATIONS
which gives rise to a relation <f>(x, y, z) = e must be such that
P, Q and R are proportional to
d<t> dtb ^ d<t>
~, ~ and ~,
dx dy dz }
respectively, and these relations cannot hold for all values of P,
Q and R.
59. To determine when an equation of the form
Pdx 4- Qdy + Rdz =
has a solution of the form <(#, y, 2) = c.
If it be assumed that Pdx -f Qdy -f Rdz = has a solution
<(#, y, 2) = c, then P, Q and R must be proportional to
respectively, or
,
^- , -5- and ,
^o; ^/ dz
where /A is a certain unknown function. From the first two of
these equations there results
or
Similarly, by using the first and third equations we get
dR dP
and by using the second and third,
SQ dB
EQUATIONS IN TWO DEPENDENT VARIABLES 81
Multiply equations (1), (2) and (3) by R, Q and P respec
tively, and add.
_ + __ = . (4)
dz/ T \dy dx ]
Therefore, if the equation Pdx -j- Qdy -|- Rdz = has a solution
<(#, y, 2) = c, equation (4) must be satisfied.
Conversely, if equation (4) is satisfied, the equation
Pdx + Qdy + Rdz =
has a solution <f>(x, y, 2) = c. The proof of this theorem is some
what long and will not be given in this book.* The theorem
however will be assumed in the subsequent work.
Definition. Equation (4) is called the condition of integra-
bility of the equation Pdx + Qdy + Rdz = 0.
60. To solve the equation Pdx + Qdy -f Rdz = when the
condition of integrability is satisfied.
Suppose at first that z is constant so that the equation becomes
Pdx 4- Qdy = 0. Solve this equation. Suppose that the solu
tion is f(x, y, 2) = a constant. Let u = f(x, y, z). Find a
quantity /A such that
du
Multiply the equation Pdx + Qdy -f Rdz = by /*.
. . ii.(Pdx + Qdy + Rdz) = 0.
This equation may be written in the form du -f Sdz = where
u and S are in general functions of x, y and z. In the equation
du -\- Sdz, change the variables from x, y and z to x, u and z by
means of the relation u = f(x, y, z). The equation then be-
* For a proof of this theorem and also that S / of Art. 60 does not con
tain x, the student is referred to Forsyth, A Treatise on Differential Equa
tions, Art. 152.
82 SHORT COURSE ON DIFFERENTIAL EQUATIONS
comes du -j- S dz = 0. It can be shown that S does not contain
x. Assuming that it does not, the equation du -\- S dz can be
integrated as an equation in u and z. The general solution of
the equation is the general solution of the original equation.
EXAMPLE. Solve the equation
yz
dx Z XZ 2 dy tan" 1 - dz = 0.
Suppose that z is constant. The equation then becomes
dx -3^ dy = or ydx xdy = 0.
2 2
, ,
x z + y a;
The solution of this equation is
- = a constant.
y
Let w = .
dx
Let uP = -.
Multiply the original equation by
Now
7 1 j ^
ef = -<iaj-
Substitute
x
u
EQUATIONS IN TWO DEPENDENT VARIABLES 83
in this equation, y being derived from the equation
x
= y
u? 4- 1 1
. . du - -2 tan l - dz = 0.
z u
Separate the variables.
du dz Q
O 2 -f man" 1 -
v u
Let tan l - = v.
u
. . vz = c.
. .z tan" 1 y - = c.
x
Therefore
z tan" 1 - = c
a;
is the general solution of the original equation.
61. Suppose that in the equation
Pdx + Qdy + Rdz =
the condition of integrability is not satisfied. Then there is no
relation <f>(x, y, z) c which satisfies the equation. In such a
case a relation
is assumed arbitrarily and a relation <(#, y, z) = c is sought
which, together with \j/(x, y, z) = 0, will satisfy the equation.
By differentiation of \j/(x, y, z) = there results
Ox " dy dz
From this equation and (1) suppose that z and dz be eliminated.
Then there will result an equation of the form P dx -f- Q dy =
84 SHORT COURSE ON DIFFERENTIAL EQUATIONS
where P and Q f are functions of x and y the values of which
depend upon $(x, y, z). Suppose that a solution of this equation
containing an arbitrary constant is found and is <(#, y, z) = c.
Then this solution and ^(#, y, 2) = together give a solution
of the equation.
As an illustration consider the following example :
The equation
di .
considered in exercises 47 to 50 inclusive, Chapter III, for special
cases of e, does not satisfy the condition of integrability if e, i
and t are variables independent of each other. For, the equa
tion may be written as
Ldi 4. (Ri - e)dt + de = 0.
By application of the condition of integrability there results
dL d(Ri - e) 1
\a*- ~di j-
or
-L = 0.
Since L is not zero, the equation does not satisfy the condition
of integrability. Assume e =/(0> however, and the equation
becomes an ordinary linear differential equation of the first order.
The solution is
-
~ -- ft
From this solution the results of exercises 47 to 50 inclusive,
Chapter III, may be found by substitution.
62. The cases considered thus far consisted of one equation in
two dependent variables. Another important class of equations
is the case of two total differential equations in two dependent
EQUATIONS IN TWO DEPENDENT VARIABLES 85
variables where each equation is of the first degree with constant
coefficients. The method of solution of this class of equations is
as follows :
By differentiation and elimination, obtain one equation in one
unknown. This equation may be solved by methods previously
discussed. The solution found must be a solution of the original
equations. Another solution is found by substituting the one
just found in the equations. The complete solution consists of
two linearly independent relations between the variables.
EXAMPLE. Solve the equations
Differentiate (2) with respect to x.
. d*y dy dz
CM b ^ +5 ^ = U
Multiply (1) by - 5, and add to (2) and (3).
.-.gZ-sf + 4y = 0.
da? dx T
This is a linear differential equation of the second order with con
stant coefficients and right hand member zero. It can therefore
be solved by the methods of Art. 31.
y = <Vf + c^. (4)
Substitute this value of y in (2) and solve for z.
--^-h jfV"- (5)
Equations (4) and (5) together constitute a set of solutions of
the given equations.
86 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXERCISES
In each of the seven following equations, show that the condi
tion of integrability is satisfied. Solve the equation.
1. (y 4 z)dx 4 (z 4 x)dy + (x + y)dz = 0.
2. (2aty 4 2^ 2 4 2xyz
3. (2a?y + 2 2 )^ 4. (^ 2
4. (a -f 2)yc?^ -f (a -f-
6. (y + 2a?)d + (xz + 2y)rfy + (xy + 2)cfo = 0.
7. (2#2/z -f ?/ 2 2 -f yf)dx 4- (^ 4 2^3 4 ^ 2 )d?/
4- (^ 4 xf 4
Solve the following sets of equations.
8. ? 4 7y - 3z = 0, 7 4 63^ _ 36. = 0.
11.
EQUATIONS IN TWO DEPENDENT VARIABLES 87
* + + *_<! *l + 3y + 4, = e >:
.*-* + *.,, * + *-._*
17.
ANSWERS
1. xy 4- yz 4- zx = c.
2. y?y 4- y*z 4- log(# -|- y 4. z) = c.
3. nfy -f ?/ 2 2 4. z*x = c. 4. xy c(a + z}.
7. xyz(x -f i/ -f z) = c.
Txi 17 fr .
8. v = c^- 32 -
9. y = Cl
10. = c
11. y = Cje 2 4. e 2 e 2
12. y = e^ x cos a; + c 2 e** sin -^- a?,
88 SHORT COURSE ON DIFFERENTIAL EQUATIONS
13. y = c^eT* cos V&E -j- c 2 e~ x sin V5#,
z = c t V5e~* sin V5# 4. c 2 V5<T Z cos V&c.
14. y = x \xe x 4- Cje" 4- c 2 e~ x ,
15. 2/ = iV^
16. y = ^ e 3 * + ^e x -f Cl e-- y ** 4- c 2 e-^ 2+ ^^,
2 = 2-V" + i 6 " + e i V3 e- (2 - y} * - c 2 V3 e
17. y = - yjg 4- A + c^ 1 -"^ 4- e
_ c 2 (3
18. y = T el ~k e ** + C i e5z cos ^5x 4. c 2 e 5a: sin V
19. y = i^e 3 * + \e^ 4 cf cos VlO^ 4- c 2 e 2x sin VI Oa?.
r
e 4a: -{- j - e 2x cos VlO^ - - e 2x sin
CHAPTER VIII
PARTIAL DIFFERENTIAL EQUATIONS.
63. So far we have considered differential equations in which
there is only one independent variable. We shall now consider
equations involving two independent variables. Such equations
belong to the class called partial differential equations,
In this book, the independent variables will be denoted by x
and y, and the dependent variable by z. The partial derivative
of z with respect to x and with respect to y will be denoted by p
and q, respectively.
Definition. A linear partial differential equation of the first
order is an equation of the form
Pp+Qq = B,
where P, Q and It are functions of x, y and z, and do not con
tain p or q.
64. If there are two equations containing x, y and z, p and q,
which can be solved for p and q, the result may be substituted in
dz = pdx -f qdy
thus giving an ordinary differential equation. Usually, however,
there is only a single differential equation given.
65. Derivation of a partial differential equation.
(a) By the elimination of constants. Let <j>(x, y, z, c v c 2 ) =0
be a relation between x, y, z and two arbitrary constants c t and
c 2 . By differentiation of <(.#, y, z, c v c 2 ) = with respect to x
holding y constant there results
90 SHORT COURSE ON DIFFERENTIAL EQUATIONS
By differentiation with respect to y holding x constant there
results
By means of these two equations and <j>(x, y, z, c v c 2 ) = 0, c v
and c 3 can be eliminated. The result is an equation
F(x, y, z, p, q) =
which is a partial differential equation of the first order.
EXAMPLE. Let x* + f -f z 2 -f c^x -f c t y = be an equation
between #, y and z, and two arbitrary constants c x and c 2 . By
differentiation with respect to x holding y constant there results
2x + GI + 2zp = 0.
By differentiation with respect to y holding x constant there
results
2y + c 2 + 2z? = 0.
By elimination of Cj and c 2 between the three equations there
results
x* -f tf z* -}- 2#zp -f 2^5 = 0.
This is a partial differential equation of the first order.
(6) By the elimination of an arbitrary function. Suppose
that u and v are functions of the variables x, y and z, and that
<(w, -y) = where <j>(u, v) is an arbitrary function of u and v.
The differential of <(>> v) = is
d<, <,
S^dtt-L -^(fo = 0.
C7M CV
Now
dw , 5t* ,
aw = ^- aa; 4- ^- dz
dx dz
when y is constant, and
, du j du ,
du = ay -)- ^- a 2
dy r ^2 y
when x is constant, and similarly for v.
PARTIAL DIFFERENTIAL EQUATIONS
91
Therefore the partial derivatives of the equation <#>(i*, v~) =
with respect to x and y, respectively, are
dd> f du du H d<b f dv dv
and
d< [" dv dv ft
Eliminate * and ^ from these equations.
^
du
du du
dv
dv dv
dv
du Su
When arranged in powers of p and q and the coefficients ex
pressed as determinants, the equation becomes
du du
dy dz
dv dv
dy dz
P +
du du
dz dx
dv dv
dz dx
,.
du du
dx dy
dv dv
dx dy
This may be written in the form
where
Pp + Qq = R
(1)
R
du du
"
du du
_
du du
dy dz
dz dx
dx dy
dv dv
dv dv
dv dv
dy dz
dz dx
dx dy
This is a partial differential equation of the first order. There
fore from the equation <j>(u, v) = a partial linear differential
equation of the first order can be formed which does not contain
the arbitrary function <j>(u, v).
EXAMPLE. Suppose that u = x -f- y -f z and v x*-\- y*-}- z 2 .
Let <(, v) = be an equation connecting u and v where
<f>(u, v) is an arbitrary function of u and v.
92 SHORT COURSE ON DIFFERENTIAL EQUATIONS
By differentiation of
respect to y there result
v) = with respect to x and with
and
respectively. By elimination of ^- and -~ from these equations
there results
1 1
1 1
C %
1 1
or
This is a partial linear differential equation of the first order
which does not contain the arbitrary function
66. We have seen that a differential equation with two inde
pendent variables can be derived from an expression containing
two arbitrary constants or from an expression containing an arbi
trary function of two independent functions of the variables. We
see therefore that a differential equation with two independent
variables may involve in its solutions, arbitrary constants or an
arbitrary function of the variables.
Definitions. A relation between the variables of a differential
equation with two independent variables which includes two arbi
trary constants is called a complete integral of the equation.
A relation between the variables of a differential equation with
two independent variables which involves an arbitrary function
of two independent functions of these variables is called a general
integral of the equation.
There is another class of solutions called singular integrals but
these will not be considered here.
PARTIAL DIFFERENTIAL EQUATIONS
93
67. Consider the two equations u c l and v = c 2 where u and
v are functions of x, y and z, and c t and c, are arbitrary con
stants. By differentiation of u = c x and v = c 2 , there result
and
du , du , du 7
- dx -f -5- d y + x- =
d# ~ dy dz
dv , dv , dv , ~
eta -f- rty + ^- dz = 0,
respectively.
Multiply (1) by ^- , (2) by ^- , and subtract.
a)
(2)
du du
du du
dz dx
dy dz
dx -
dv dv
dv dv
dz dx
dy dz
Multiply (1) by ^-, (2) by^-, and subtract.
du du
du du
~dx fty
dy dz
dx-
dz = 0.
dv dv
dv dv
dx dy
dy dz
dx dy dz
du du
du du
__
du du
dy dz
dz dx
dx dy
dv dv
dv dv
dv dv
dy dz
dz dx
dx dy
Now
if
v) is a general integral of the equation
Pp + Qi = R,
Q R
du du
du du
du du
dy dz
dz dx
dx dy
dv dv
dv dv
dv dv
dy dz
dz dx
dx dy
See Art. 65.
94 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Therefore <(, v) = is a general integral of the equation
Pp -f- Qq = H if u = c l and v = c 2 are solutions of the equations
68. From the investigations of Arts. 65 and 67 the following
rule for finding a general integral of the linear partial differential
equation Pp -f- Qq E is determined.
Solve the equations
dx dy dz
Suppose that u = ^ and v = c 2 are two independent integrals of
these equations. Then <(>, v) = where <O, v) is an arbi
trary function of u and v is a general integral of the equation
= R.
Definition. The equations
dx dy dz
are called the subsidiary equations of Pp -f Qq = E. They are
also sometimes called Lagrange s equations.
69. As illustrations of the method of solution of a linear par
tial differential equation of the first order, consider the following
examples.
EXAMPLE 1. Solve the equation x*p -}- xyq -f- y* = 0.
Write the subsidiary equations
dx dy dz
~tf = xy = ~ f
Solve the equation
dx dy x
x* xy y
Solve the equation
dy dz
PARTIAL DIFFERENTIAL EQUATIONS 95
From
x
-,-,
substitute the value of x, and the equation becomes
A general integral of the original equation is therefore
EXAMPLE 2. Solve the equation (y z}p-\-(z
Write the subsidiary equations
dx dy dz
y-z~z-x~x-y
From a familiar theorem of algebra, if
then la -f me -f ne = Ib -f md -f- n/ where , m and w are any
multipliers whatsoever. Application of this theorem to the sub
sidiary equations gives
dx -f dy -f dz = 0, (1)
when I = m = n, and
-f ydy + zdz = 0, (2)
when I = x, m = y, n = z.
Solve equations (1) and (2). Therefore x + y -f z = c l and
x z -\- y* 4- 2 2 = c 2 are solutions of equations (1) and (2), and
therefore of the subsidiary equations. A general integral of the
original equation is therefore <j>(x -f y -f z, x 2 -{- i/ 2 -f 2 2 ) =0.
96 SHORT COURSE ON DIFFERENTIAL EQUATIONS
EXERCISES
Determine the partial differential equations of which the four
following equations are complete solutions, c x and c 2 being arbi
trary constants.
1. z = CjX + c$. 2. z 2 = c^ 2 + c 2 </ 2 .
X* V* Z 2
3. z = O + Cl )(y + c 2 ). 4. -i+ jr+ - = !
C l C 2
Eliminate the arbitrary function from each of the four follow
ing equations.
5. <j>(x+y-z, z 2 +*/ 2 -z 2 )=0. 6. <fr(x + y + z, z) = 0.
7. = 6*K* + y). 8- *=/(* 2 + 2/ 2 ).
Find a general integral of each of the following equations.
9. xzp yzq = xy. 10. x*p + y*q z 2 = 0.
11. a?yp -f- yq = x*z. 12. xp yq = a? ?/.
13. (^ - z 2 )^ + (z 2 _ x^q + (^ - * 2 ) = 0.
14. (2z _ 3t/)j9 + (3a? - 4z)q = 4y - 2x.
ANSWERS
1. xp -|- yq = z. 2. xp -\-yq = z.
3. jo^ = z. 4. xzp -}- 2/z# z 2 -f a 2 = 0.
5. (yz)p+(zx}q=yx. 6. p q = Q.
7. p q = z. S. yp xq = 0.
9. <f> ( #v, log w -{- ^ )=0. 10. < ( - -, - - ) =0.
12. <A(^, a: + y - z) = 0.
13. ^ + y + z, a; 3 + f + z 3 ) = 0.
14. <i>(4x + 2y + 3z, ^ 2 -f 2/ 2 + ^ 2 ) = 0.
CHAPTER IX
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS.
INTEGRATION IN SERIES
APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
70. It is shown in the Analytical Theory of Heat that the
change of temperature in any solid at a point (#, y, z) within
the solid is given by the equation
where u represents the temperature at the point and t denotes
time.
In polar or spherical coordinates the equation becomes
^ r -
du
__
dt ~ r 2 ^^ sin 00
and in cylindrical coordinates,
du A d*u 1 5w 1 d*u a*Ml
c5^ = L 5r 2 + r dr + r 2 a> T + d2 \
If the solid is a rectangular plate so thin that the thickness
need not be taken into account, equation (1) becomes
du ran, ann
= c " +
If the solid is a wire of infinite extent so thin that the breadth
or thickness need not be taken into account, equation (1)
becomes
du d*u
~dt= c W
In the case of a sphere when the temperature u depends merely
on the distance of the point from the center, equation (1), as
8 97
98 SHORT COURSE ON DIFFERENTIAL EQUATIONS
can be seen from (2), reduces to
d(ru) a (ru)
~df~
In the problem of permanent states of temperature, du/dt = Q,
and the equation becomes
an equation known as Laplace s Equation, and sometimes writ
ten V 2 u = 0. This equation also figures in the Theory of
Potential.
In polar or spherical coordinates, equation (7) becomes the
right-hand member of equation (2) set equal to zero, and in
cylindrical coordinates, it becomes the right-hand member of
(3) set equal to zero.
In the Theory of Acoustics, in considering for instance the
transverse vibrations of a stretched elastic string, there occurs
the equation
&y_^y m
Bf dx"
and if the resistance of the air be taken into account, the
equation
In the problem of the vibrations of a stretched elastic mem
brane, there occurs the equation
which in cylindrical coordinates becomes
2 ^J
71. As an illustration of transformation of coordinates, con
sider the transformation of Laplace s Equation in two dimen
sions,
APPLICATIONS OF EQUATIONS 99
dx* cty 2 =
from rectangular to polar coordinates.
The equations of transformation are x = r cos 6, y = r sin 6.
Now u is a function of x and y and therefore of r and 0.
Therefore, as seen in calculus,
. du , du
du = dr -f- ~ dd.
dr r dO
If y is held constant, this equation becomes,
du 7 du , ,
Divide by A#, or what is the same, dx, and there results th&
equation
du du dr du dO .p
dx ~~ dr dx dO dx *
Similarly,
du du dr du dO ,~,
Since x = r cos and y = r sin 0, therefore r = V# 2 + y* and
= tan" 1 y/x.
dr x x .
- fl~ = = == - = cos 0,
dO y y sin &
and
dO x x cos
Now
du du dr du dO du du sin
a~ = * 5~ + HZ a- = a~ COS ^ J^fi >
da; c?r do; 50 ^* dr du r
and
100 SHORT COURSE ON DIFFERENTIAL EQUATIONS
d*u d\du dusinOl d [du du sin 0"] sin b
^~a = 5~l a" COS0 -=-- COS 6 ^7. =- COS0 * - -
dx* dr\dr dO r \ dO\_dr dO r j r
d*u sin0 dusinO ]
cos -^^ - + -^ T- cos B
drdO r ~ DO r 2 ,
ducosOlsinO
- ae ~r \ ~
[du .
- s~r s
Similarly
a 2 i* [ d\i . d u cos du cos 6 "I .
a~i = ^^ Sm ^ + a 3d ~ ^7 s- sm ^
^2 |_ar 2 ^ drdO r dO r 2 J
d 2 ^ . .. ait .,, a 2 it cos ait sin 0~| cos
Lara0 aV a0 2 r a0 r r
a 2 it a 2 w a 2 it i ait i a 2 i*
a^c 2 a^/ 2 a/* 2 T* a?* y 2 a0 2 *
Therefore the equation
a it a u
ox uy
in polar becomes
_ i . i_ _ u _ o.
^2 1^ y. ^^ \ y,2 ^Qt
72. A method of determining particular solutions of those of
the above equations with constant coefficients is illustrated in the
following example.
EXAMPLE. Find particular solutions of the equation
z z d z z
Assume that there is a particular solution in the form
z = e^+fly+Y* where a, ft and y, are constants.
Substitute in the equation.
. . y 2 e*+0y+Y = C 2 (a 2 -f p^e^+M+y .
Now e^+^y+v cannot be zero for any values of x, y and t.
APPLICATIONS OF EQUATIONS 101
Therefore z = e^+Pv^^+P 2 is a particular solution of the
equation where a and ft are arbitrary constants.
The above solution can be put into another form as follows:
Let a = aj and (3 = /3j where j = V 1.
Then
z =
Therefore
z = sin (ax -f fty ct Va 2 -f ff), (1)
and
3 = cos (ax + (3y+: ct Va 2 -f 2 ), (See Art. 5. ) (2)
are particular solutions of the equation.
\ From these can be found particular solutions in the forms
z = sin ax sin fiy sin ct Va 2 -f-
z = sin ax sin
and six others. The determination of these six is left as an
exercise to the student.
73. Consider the equation
+ -. :
dr
which is Laplace s Equation in spherical coordinates where u is
independent of <.
Let u = r m P, where P is a function of alone, and m is a
positive integer. On substitution there results the equation
m(m
sin dO
Change the independent variable from to x where x = cos 0.
0. (2)
The solutions of equation (1) are known when P is determined
from equation (2). Equation (2), not only when m is a posi-
102 SHORT COURSE ON DIFFERENTIAL EQUATIONS
tive integer but for all values of m, is called Legendre s Equa
tion. Its solutions are discussed in Arts. 76, 85 and 86.
74. To find particular solutions of the equation
which is equation (11), Art. 70, when z is independent of </>,
let z = R - T where R is a function of r alone and T is a func
tion of t alone. Substitute in the equation.
^d^T
or
^TW = R \W~t~r drj" w
The right-hand member of (2) does not involve t. Therefore
the left-hand member does not. The left-hand member does not
involve r. Therefore the right-hand member does not. There
fore each member is constant. Call the constant /u, 2 .
and
1 dR
A particular solution of (1) is therefore R T where T is de
termined by equation (3) and R by (4).
Particular solutions of equation (3) are T = cos pet and
T= siufjict. (See Art. 31.)
To solve equation (4), let r = x/p and substitute in the
equation.
...g+lg+J^O. (5)
dx* ~ x dx ^
Equation (5) is a special case of the more general equation
INTEGRATION IN SEEIES 103
known as Bessel s Equation. Its solutions are considered in
Arts. 79 to 84 inclusive.
INTEGRATION IN SERIES
75. It will be noted that as yet in this book no equations with
variable coefficients, of higher order than the first, have been
considered except a few very special cases discussed in Chaps. V
and VI. The remainder of this chapter is devoted to a discus
sion of linear differential equations of the second order with coef
ficients rational integral functions of x, and second member zero.
To such a set belong Legendre s and Bessel s Equations men
tioned above.
Not all differential equations, not even all in the comparatively
simple form of linear differential equations of the first order, are
capable of solution in finite form. When solutions cannot be
found in finite form, recourse is had to integration in series. In
the set about to be considered, some equations have solutions in
finite form and some have not.
We shall attempt here to find solutions only in the form of
infinite, convergent, power series.
If an equation be capable of solution in finite form, this form
is found when a solution is attempted in the form of a power series.
For instance, in exercise 11, page 120, the solution found as if it
were made up of infinite series is in reality in finite form.
Sometimes the series that make up the solution of an equation
may be recognized as those of familiar functions. In such cases,
the solution can be written in terms of those functions. For
instance, in the answer given on page 122 for exercise 12, page
120, if J. be taken equal to 2 and B to 1, and the two particular
solutions be added, there results the series
which is x~ y e* x . If the second solution be subtracted from the
first, there results the series which is x~ 2 e~ 2x . The general solu
tion is therefore
104 SHORT COURSE ON DIFFERENTIAL EQUATIONS
y = CjOTV* -|- c 2 3T 2 e~ 2z .
76. Let us attempt to find a solution of Legendre s Equation
in the form of a power series in x.
At first, assume that there is a power series
V=00
y == ff(fi ~\~ o \% K ~f~ * * * ~i~ g * -i- * * * x O V X K
where g , g v - , K are constants, which will formally, i. e.,
without regard to whether the series converges or not, satisfy the
equation. It is no restriction to assume, as we shall, that # =j=0,
because, if there is any solution at all, one at least of the g s is
not zero, and we assume that the series begins with the term con
taining the first g which does not vanish.
Since
I/--0
and
^ = S> (/< + v)(/c
Substitute in the equation.
V 1Y1 r Vir _i_ i/V* _i_ !
. . 2^ lA 1 x ) \. K H- V )(, K + v
or
j/=a
!/=(
If
- m(m + l)}flr^+"] = 0.
INTEGRATION IN SERIES 105
is to satisfy the equation, the coefficients of each power of x in
(1) must be zero. Therefore there results the following series
of equations :
(K + 2)(K + 1)0, - {K(K + 1) - m(m + 1)}^ = 0,
3)(K + 2)</ s - {(K + 1)(* + 2) _ m(m + 1)}^ = 0,
(2)
2r - 2)(/c + 2r - 1) - m(m
From the first of these equations, since g =j= 0, therefore,
= 0, or K = 1. At first, take K = 1.
Substitute in equation (2) and calculate the # s in succession.
ffi =
X
1
where ^ is arbitrary, and g zr has the value given above, formally
satisfies the equation. Since this series is convergent, (3) is a
particular solution of the equation.
Next, take K = 0.
Substitute in equations (2). g l is arbitrary. Call it zero.
106 SHORT COURSE ON DIFFERENTIAL EQUATIONS
m(m+l) m(m-2)(m + l)(
" ~
^ l - 4
[2 ~U~
(4)
where <7 is arbitrary, and g 2r has the value given above, formally
satisfies the equation. Since the series is convergent, (4) is a
particular solution of the equation.
If solution (3) be denoted by y l and solution (4) by y v the
general solution of the equation is y = Ay l -f By^ where A and B
are arbitrary constants. (See Art. 11.)
77. The general form of a linear differential equation of the
second order with right hand member zero is
It will be assumed here that q Q (x), ?,(), g 2 () are rational
integral functions of x.
If a solution is to be found in the form of a power series in
x a, it will be convenient to write the equation in the form
(* - )>(*) g + (X - ) A 00 g + A ( Z ) . y = 0, (2)
where .p (a;), ^(a;), jt> 2 () are rational integral functions of x.
The equation can be written in this form in an unlimited number
of ways by multiplying it through by a suitable power of x a
and a rational fraction neither the numerator nor denominator of
which contains x a.
INTEGRATION IN SERIES 107
Definition. The point a is a regular point of equation (2) if
p.W * o.
Without at first making any assumption with regard to the
point a, substitute
in equation (2) and attempt to determine the g s so that the
equation is formally satisfied.
= 0. (3)
Call the expression in square brackets f(x, K -f- v).
Develop f(x, K -j- v) into a power series in x a by Taylor s
Theorem.
.-./*, K + V) =/(a, K + V) + / (, K + V) ~~ +
Substitute this development in (3), equate each power of
x a to zero and there results the following series of equations:
<7j(a, K + l) +flr / (o, K) = 0,
2) + ^/ (a, K + 1) +
Now g 4= 0. Therefore /(a, K) = 0.
And /(a, K) = K(K -
108 SHORT COURSE ON DIFFERENTIAL EQUATIONS
.-.K(ic-.l)^ (a) H-icp^a) + ^(a) =0.
From this equation can be determined the value or values of K
which are to be used in the subsequent equations (4). If p (a}
is not zero, the equation is of the second degree. If p (a) is
zero, the equation is of lower degree than the second.
The necessary and sufficient condition that the equation
is of the second degree is therefore that the point a be a regular
point of the differential equation.
Definition. The equation K(K l)j (a)-f ^(a) + p 2 (a) =
is called the indicial equation of the differential equation (2).
If the point a is regular, the indicial equation gives two values
of /c, say K and K", and from equations (4), for either value of
K, the values of g v g 3 , , may be computed, in general, in
terms of g .
Therefore in general there are two series in ascending powers
of x a, namely,
2/ = V Z U,(z-a>y +v ,
v=0
and
y-Y* (*-)*"*,
v=0
where g is arbitrary in either series, which will formally satisfy
equation (2).
78. The following theorems with regard to the solutions of the
differential equation
in a power series in x a have been established. The proofs
are too long to be given in this book. For a discussion of these
theorems the student is referred to a pamphlet entitled Regu
lar Points of Linear Differential Equations of the Second Order "
by Professor Maxime Bocher, published by Harvard University.
INTEGRATION IN SERIES 109
Theorem I. If a is a regular point of the differential equa
tion, and the difference of the roots of the indicial equation is not
zero or a positive integer, two solutions in the form of a power
series in x a, viz.,
and
y = E </ v (x-a>y +v ,
v=0
where K and K" are the roots of the indicial equation, exist, and
these series are convergent. In each of these series g is
arbitrary.
In this case, if y l denotes one of the series and y 2 the other,
the general solution of the equation is y = Ay^ -j- Ey % where A
and B are arbitrary constants.
A case to which this theorem applies is Bessel s equation when
n is not zero nor an integer, discussed in Art. 80.
Note. By the difference of the roots of the indicial equation
being a positive integer is meant that the greater minus the less
is a positive integer.
Theorem II. If a is a regular point of the equation and the
difference of the roots of the indicial equation is a positive inte
ger n, the necessary and sufficient condition that two solutions
of the form under Theorem I exist is that
?-,/ (, "" + - 1) + + 9. f (a ~^ = 0,
IV
(see equations (4), Art. 77), where K" is the smaller of the
roots, and when this condition is fulfilled, the series are conver
gent,
In this case the series corresponding to the larger value of K,
say K , can be found as before. In the series corresponding to
K", g and </, are arbitrary, but if g l be chosen zero a particular
solution in terms of g is found. Then if y l denotes the first
series, and y 3 the second, the general solution of the equation is
y = Ay r -j- By t where A and B are arbitrary constants.
110 SHORT COURSE ON DIFFERENTIAL EQUATIONS
A case to which this theorem applies is Legendre s Equation
discussed in Art. 76. ,
Theorem III. If a is a regular point of the equation, and the
difference of the roots of the indicial equation is either zero, or a
positive integer n where
two solutions are found, one being
y = T,9 v (x-aY +v ,
v<d
the other being
y = log O _ a) fr(* - a Y +v +"if0r(* - a)""*",
v=0 v=0
where K is the larger root of the indicial equation and K" the
smaller, and these series are convergent. In the first of these
series g Q is arbitrary. In the second, g Q is arbitrary and g v is
determined in terms of g .
If y l denotes the first series and y 3 the last; term of the second,
the general solution of the equation is
y= 1 A + B lo g O - 01 2/1 + By
where A and B are arbitrary constants.
A case to which this theorem applies is BessePs Equation
when n = 0, or an integer, discussed in Arts. 82 and 83.
Theorem IV. If the point a is not a regular point of the
equation there are not two solutions of the equation in any of
the forms under Theorems I and III, and if any series in one of
these forms is found it is usually not convergent.
Cases to which this theorem would apply will not be considered
in this book.
BESSEL S EQUATION
79. We shall now consider the solutions of the equation
in the form of a power series in x.
INTEGRATION IN SERIES 111
The equation as it stands is in the form (2) of Art. 77, where
PoW = * PM = ! ^aW = x * - n *-
Since jp (#) cannot be zero for any value of x, all points of
this equation are regular. Therefore the solutions of the equa
tion for all values of x, and in particular when x = 0, will come
under one or other of the forms mentioned in the first three
theorems of Art. 78.
Substitute
y = " g v ^ v
i>=0
in the equation.
- "If [(* + V)(K + v - 1) + -f v) + (V _ 7i)]^^+" = 0,
v=0
or
If [(K + v) 2 + * 2 _ ^ 2 ]^^^" = 0.
x=0
The equations for the determination of the g s are therefore :
(K 2 _ n^g. = 0,
[(*+!)* -0^ = 0,
[(x + 2) 2 -n 2 ]^ 2 + !7o = 0,
[(K + 3) 2 _71 2 ]^ = :0, (1)
[(K + 2r - I) 2 - <]0 M = 0,
Since gr =j= 0, from the first equation there results K = n.
The difference of the roots of the indicial equation is therefore
2n. If n = 0, this difference is zero. If n = ~, where j9
A
is an odd integer, or if n is an integer, this difference is a posi-
P
tive integer. If n is neither zero nor ^ nor an integer, the dif-
Zi
ference is neither zero nor a positive integer.
nr\
80. At first assume n neither zero nor ^ nor an integer.
112 SHORT COURSE ON DIFFERENTIAL EQUATIONS
This is the case covered by Theorem I. There are therefore
two solutions
and
To determine the g v substitute K = n in equations (1).
<7o
4(2 + 4) ~ 2 4(2ra + 2) (2m + 4)
_ f _ 1 y 9
-^ >
2 - 4 - - - 2r(2 + 2)(2n + 4) - - (2w + 2r)
2 4(2 + 2)(2n + 4) ~ g
where g is arbitrary and ^ 2r has the value given above is a par
ticular solution of the equation.
Similarly, on substituting K = n in equations (1) there
results
y = gf~ [l + 2(2n g _ 2) ,
_ X __ I . . 1 ^Zf^r , . /Q\
f 2.4(2n_2)(2n_4)" f ^ g J Lj
where ^ is arbitrary and
9,
_____ _
~ 2 4 - 2r(2^i - 2)(2w - 4) (2 - 2r)
is a particular solution of the equation.
INTEGRATION IN SERIES 113
If y l denotes the first series and y t the second, the general solu
tion of the equation is y = Ay 1 -f- By 2 where A and B are arbi
trary constants.
79
81. Next, assume n = =^^- Assume, for definiteness, that p
is positive. In this case the difference of the roots of the indicial
equation is a positive integer, viz. p. From an examination of
T)
equations (1) it is seen when n = ^ that both g and g p are
2i
arbitrary. Choose g p = 0, and there results the same equation
nr\
as (3) of the preceding article when -j-^ is substituted for n.
Therefore in this case there are two particular solutions of the
equation which are the same as the solutions in the case of the
f)
preceding article when ^ is substituted for n.
Zi
82. Next, assume n an integer. Since n appears only in the
form of a square in the differential equation, it is sufficient to
suppose it a positive integer.
In this case the difference of the roots of the indicial equation
is the positive integer 2n.
For the root K = n, the series is the same as (2) of Art. 80.
For the root K" = n, the equation
is such that the coefficient of g yn is zero. Therefore, since
#2n-2 =H 0> tms case comes under that mentioned in Theorem III,
Art. 78.
To get a solution corresponding to K", let
y = qga
i/=0 v=0
For the purpose of determining the coefficients ~g v , write the
series in the form
y = if (v-2n log X + g v
v=0
where g_ tn = g_, n+l = = g_, = 0.
9
114 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Substitute in the equation.
-"if [(*" + ") 2 + *? - <l#v-2 log x x*"+ v
"V{2(K"+v)^_ 2 ,+ [(K-- f v) 2 +^-^]^}^+" = 0. (1)
Now in the coefficients of log x - X K "+ V , <7_ 2n , g_ 2n+l , , g_ v
are zero, and the remaining ones are the same as the left hand
members of equations (1), Art. 79, with K" -{- In substituted for
K. Now K" -{- 2n = K , and equations (1), Art. 79, hold for
K K . Therefore all the coefficients of log x - X K " +V vanish.
From the second set of terms in (1) are found the equations
from which to determine </. These equations are:
<V 2 _ n^g. = 0,
[(K- + 2ny - n*]g^ + g M + 2(>" + 2n) 9o = 0,
[(K" + 2n H- I) 2 - n^g, n+l + 2(" + 2n + 1)^ = 0,
" + 2n + 2) 2 _ n^g M + 9*> + W + 2w + 2)(/ 2 = 0,
Since in these equations, K" = n, therefore g , an arbitrary
constant, satisfies the first equation. Also,
_ 2)
- 4) *
r 2.4. .-2r(2n-2)(2n_4)- - -(2/i-r
Since the coefficient of <? 2n is zero, this equation introduces no
new g. The equation, however, gives a means of determining the
hitherto undetermined constant g ,
INTEGRATION IN SERIES 115
Also
~2 2 "- 2 Q-l) 2
~fi 7, - ff 77 _ o _
<Ji - > U* - 2(2n - 2) J * n ~* -{2.4.6-. -~(2n - 2) } 2
where
Choose </ 2n = 0. Therefore
-| -v y J
; 2 4 ... 2r2^
z 4
2) 2 2n
^ 1
+ 2 J
r n+2r ,
where ^r is arbitrary and <7 2n+2r has the value given above, is a
particular solution of the equation
If the first solution be denoted by y^ and all terms not involv
ing log x in the second by ?/ 2 , the general solution of the equation
is y = ( A -\- B log x)y l -\- JBy z where A and B are arbitrary
constants.
116 SHORT COURSE ON DIFFERENTIAL EQUATIONS
83. Next, assume n = 0. In this case the difference of the
roots of the indicial equation is zero. This case is covered by
Theorem III.
The two infinite series when n = can be found from equa
tions derived as in the preceding case. They can also be found
by letting n be zero in the results in that case. Two particular
solutions are
z x 4 " x zr
~ 2 2 ~~ " ~^~ ^ ~ ) 2
4 2 (2r)
and
84. As will appear in applications to physical problems, when
n is a positive integer it is convenient to take, not y^ and y v but
the quotients of these by 2 n \ n where g is unity. These special
solutions are written J n (x) and W n (x) so that
~
and
f 2 n ~ 3 1 n 2 x~
F n < = J log x - 2- 1 \n - 1 x- + -
2 n ~ l \n-l
LEGENDRE S EQUATION
85. Returning now to the equation
(1 - *) g - 2*g + (w+l)y = 0,
considered in Art. 76, we see that it can be transformed into the
INTEGRATION IN SERIES 117
form (2) of Art. 77 by multiplying through by x 2 . The trans
formed equation is
* 2(i - * 2) 2 - 2x * fx + m(m + ixy =
where p Q (x) = 1 x\ p v (x) = 2x*, p t (x) = m(m -f- 1)# 2 .
Since _p (#) = when x = I and x = 1, the points 1 and
1 are not regular points of the equation. All other points are
regular.
The indicial equation when x = is
K(K_ 1) = 0.
This equation has the roots and 1. The differential equa
tion in this case comes under the case mentioned in Theorem II.
This case was already discussed in Art. 76.
86. It is convenient when m is a positive integer as it was in
the illustration of Art. 73 to have a solution of Legendre s Equa
tion as a series in descending powers of x. In this case we shall
take the equation, as in Art. 73, in the form
Let
(1 _ * ) _ 2x + m(m + 1)P= 0.
i/=0
and substitute in the equation.
"Z O - ") (n - v - I)?,**- 1 - 2
v=0
_ \_( n v)(n v -f 1) m(m -f \y\g v x n ~ v = 0.
. . [n(n+ 1) _m(m + l)]0r = 0,
\_(n - 1> _ m(m + 1)]^ = 0,
n(n - IX - [(^ - 2;(w - 1) - m(m + 1)]^ = 0,
_ [(n - 2r)<> _ 2r + 1) - m(m + 1)]^ = 0.
118 SHORT COURSE ON DIFFERENTIAL EQUATIONS
From the first equation, since g 4= 0,
. . n(n -f- 1) m(m + 1) =0.
. . n = m or n m 1.
"it first take n = m.
(m l)m
- . 9l = 0, g % = - - g v <7 3 = 0, . -,
= (~ IV ~ - - 2r - 2)
J 2r . . . 4 - 2(2m - 2r + 1) -
A series in descending powers of x which satisfies the equation
is therefore
P-
~
\l O- 1 ) -
~ 2C2^TT) x
(m-3)(m-2)(m-l)m g 1
4.2(2m_3)(2m_l) h ~g~ +
^nere (/ is arbitrary and # 2r has the value given above.
By taking n = _ m 1, there results the solution
-
f (
I
J
where ^ is arbitrary and
(m + 2r)(m-f 2r - 1) (m
2r 2r 2(2m 4. 3) - - - (2m + 2r +T) v
When m is a positive integer, solutions (3) and (4) of Art.
76 is a finite series according as m is even or odd, and in either
case, equation (1) above is finite differing from (3) or (4) of
Art. 76 only by a constant factor.
87. If series (1) of Art. 86 be multiplied by
(2m- l)(2m_3) ! 2m .
-- ^ - or , the resulting integral is
, m 2 m (m) 2
INTEGRATION IN SERIES 119
called the Legendrian Coefficient of the wth order, and is de
noted by P m (x).
The successive values of jP TO (#) are readily found to be
EXERCISES
1. Find the remaining six particular solutions of the equation
considered in Art. 72.
2. Find two particular solutions of the equation
. du ^d 2 u
dt = dx*
3. Find two particular solutions of the equation
d(ru) ,d\ru)
~ ~~
4. Find four particular solutions of the equation
du 2 / d 2 u d 2 u \
dt ~ J \ dx 2 dy 2 /
5. Find four particular solutions of the equation
_
dt 2 ~ dtf
6. Find four particular solutions of the equation
d y + U dy c> d * y .
at* + . af." dx*
7. Show that equation (1) of Art. 70, in rectangular coor
dinates, becomes equation (2) when transformed to polar or
spherical coordinates. The equations of transformation are
x = r cos sin <, y = r sin sin <, z = r cos <.
du du dr du dO du d<j> n i
Suggestion. =- = =- - + ^, -_- + ^ ^-, and similarly
dx dr dx ~ d$ dx d<$> dx
for y and z.
120 SHORT COURSE ON DIFFERENTIAL EQUATIONS
Find the general solutions of the following equations :
8. 2*-g _*!+(! +). 0.
9 " 9af &+&*-*)&- (* + V9 = 0-
10. 4z4* 1
14. x_
17. Show that,
ANSWERS
1. 2 = sin ax cos % sin at Va 2 + /8 2 ,
z = sin CUE cos fiy cos cf V 2 + ft*,
z = cos a# sin fiy sin c^ Va 2 + ft*,
z = cos ax sin /??/ cos c^ a 2 + ft 2 ,
z = cos ax cos /&/ sin ct Va 2 -f y8 2 ,
2 = COS cue COS y8y COS ct Va 2 -f y8 2 .
2. u =e ~ c2a2t cos a^, it = e- c2 2< sin
INTEGRATION IN SERIES 121
3. u = - e~ c2a2 cos ar, u = - e~ c2a2 sin ar.
r r
4. u = e -e 2 (<* 2 +0 2 )* cos ax cos fiy, u = e- c2 ( 2 +0 2 ) cos ax sin $/,
w = e -c 2 (a2+02)f g n aa , g j n ^ M _ e _ C 2( a 2+02)f g j n ax CO6 ^
5. y = cos ax cos Ca ^> y = s ^ n a:c s ^ n Ca ^f
y = sin cue cos ca^, 2/ = cos ax s ^ n Ca ^-
6. y = e~ kt sin aa; cos f cV ^ 2 , ?/ =
y = e cos cue cos f Vc 2 a 2 k 2 , y = e kt cos cue sin t VcV 2 k 2 .
8. =
1 3
274^3^7 ~ 2.4.6.3. 7-11
where
e 2 1
1 x f g ** \
and
9.
- 2-5 . 2-5-8
x-+^+^ +
where
r) _
~
~ 3 2r 5 - - - (2r + 3r 2 ) r ~ 3 2r - 1 - (3r 2 _ 2r)
10. -
122 SHORT COURSE ON DIFFERENTIAL EQUATIONS
where
_ 1 1
^ = 8 r 3 1 (2r 2 + r) ^ 2r = V- 1 - 6 (2r 2 _7)
11. =
12. y = 4s- 1 [l + f* 2 + g^g* 4
+ - + ^- + -..
+ 1^ + 2^2^ + H- g^ + ],
where
r 6-20---2(2r 2 +r)
13. .
^ 2r =
l + K + gT 30 ^ + + g
where
^ =
3r = 12.42--.3(3r 2 + r) = 6 30 3(3r 2 - r)
14. y = ^[1 - %x 4. ^ 2 ] 4. ^- 2 [l - 4a;].
15. y=[A + Blogx]
[2 2 2 2 s
1 - p + pTI ^ ~ FT2 2 T
+ B p* - j-Ai + i) + irrra-^P + 4 +
where
Or
and ..
INTEGRATION IN SERIES 123
16. =
*) + Al + 4
where
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