Power series methods of solving ordinary differential equations 505
Substitutingyandy′′into each term of the given
equationy′′− 2 y=0gives:y′′− 2 y=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1
+[a 2 (c+ 1 )(c+ 2 )− 2 a 0 ]xc+···
+[ar+ 2 (c+r+ 1 )(c+r+ 2 )
− 2 ar]xc+r+···= 0 (30)(iv) Theindicial equationis obtained by equating
the coefficient of the lowest power ofxto zero.
Hence, a 0 c(c− 1 )=0 from which,c= 0 or
c= 1 sincea 0 = 0
For the term inxc−^1 ,i.e.a 1 c(c+ 1 )= 0
Withc= 1 ,a 1 = 0 ;however,whenc= 0 ,a 1 is
indeterminate, since any value ofa 1 combined
with the zero value ofcwould make the product
zero.
For the term inxc,
a 2 (c+ 1 )(c+ 2 )− 2 a 0 =0 from which,a 2 =2 a 0
(c+ 1 )(c+ 2 )(31)For the term inxc+r,ar+ 2 (c+r+ 1 )(c+r+ 2 )− 2 ar= 0from which,ar+ 2 =2 ar
(c+r+ 1 )(c+r+ 2 )(32)(a)Whenc=0:a 1 is indeterminate, and from
equation (31)a 2 =
2 a 0
( 1 × 2 )=
2 a 0
2!In general,ar+ 2 =2 ar
(r+ 1 )(r+ 2 )andwhenr=1,a 3 =2 a 1
( 2 × 3 )=2 a 1
( 1 × 2 × 3 )=2 a 1
3!
whenr=2,a 4 =2 a 2
3 × 4=4 a 0
4!Hence,y=x^0{
a 0 +a 1 x+2 a 0
2!x^2 +2 a 1
3!x^3+4 a 0
4!x^4 + ···}from equation (28)=a 0{
1 +2 x^2
2!+4 x^4
4!+···}+a 1{
x+
2 x^3
3!+
4 x^5
5!+···}Since a 0 and a 1 are arbitrary constants
depending on boundary conditions, leta 0 =P
anda 1 =Q,then:y=P{
1 +2 x^2
2!+4 x^4
4!+···}+Q{
x+2 x^3
3!+4 x^5
5!+···}
(33)(b)Whenc=1: a 1 =0,andfromequation(31),a 2 =2 a 0
( 2 × 3 )=2 a 0
3!Since c=1, ar+ 2 =2 ar
(c+r+ 1 )(c+r+ 2 )
=2 ar
(r+ 2 )(r+ 3 )
from equation (32) and whenr=1,a 3 =2 a 1
( 3 × 4 )=0sincea 1 = 0whenr=2,a 4 =2 a 2
( 4 × 5 )=2
( 4 × 5 )×2 a 0
3!=4 a 0
5!
whenr=3,a 5 =
2 a 3
( 5 × 6 )= 0Hence, whenc= 1 ,y=x^1{
a 0 +2 a 0
3!x^2 +4 a 0
5!x^4 +···}from equation (28)i.e. y=a 0{
x+2 x^3
3!+4 x^5
5!+...}Again,a 0 is an arbitrary constant; leta 0 =K,then y=K{
x+2 x^3
3!+4 x^5
5!+···}However, this latter solution is not a separate solution,
for it is the same form as the second series in equation
(33). Hence, equation (33) with its two arbitrary con-
stantsPandQgives the general solution.Thisis always