Power series methods of solving ordinary differential equations 503
−a 0 xc+^1 −a 1 xc+^2 −a 2 xc+^3
−a 3 xc+^4 −···−arxc+r+^1 −··· (c)(iv) Theindicial equation, which is obtained by
equating the coefficient of the lowest power of
xto zero, gives the value(s) ofc. Equating the
total coefficients ofxc(from equations (a) to (c))
to zero gives:
2 a 0 c(c− 1 )−a 0 c+a 0 = 0
i.e. a 0 [2c(c− 1 )−c+1]= 0
i.e. a 0 [2c^2 − 2 c−c+1]= 0
i.e. a 0 [2c^2 − 3 c+1] = 0
i.e. a 0 [( 2 c− 1 )(c− 1 )]= 0
from which, c= 1 orc=1
2
The coefficient of the general term, i.e.xc+r,
gives (from equations (a) to (c)):2 ar(c+r− 1 )(c+r)−ar(c+r)
+ar−ar− 1 = 0from which,ar[2(c+r− 1 )(c+r)−(c+r)+1]=ar− 1andar=ar− 1
2 (c+r− 1 )(c+r)−(c+r)+ 1(25)(a) Withc= 1 ,ar=ar− 1
2 (r)( 1 +r)−( 1 +r)+ 1=ar− 1
2 r+ 2 r^2 − 1 −r+ 1
=ar− 1
2 r^2 +r=ar− 1
r(2r+1)
Thus, whenr=1,a 1 =a 0
1 ( 2 + 1 )=a 0
1 × 3
whenr=2,a 2 =a 1
2 ( 4 + 1 )=a 1
( 2 × 5 )=a 0
( 1 × 3 )( 2 × 5 )ora 0
( 1 × 2 )×( 3 × 5 )
whenr=3,a 3 =a 2
3 ( 6 + 1 )=a 2
3 × 7=a 0
( 1 × 2 × 3 )×( 3 × 5 × 7 )whenr=4,a 4 =a 3
4 ( 8 + 1 )=a 3
4 × 9=a 0
( 1 × 2 × 3 × 4 )×( 3 × 5 × 7 × 9 )
andsoon.From equation (23), the trial solution was:y=xc{
a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···
+arxr+···}Substitutingc=1 and the above values ofa 1 ,a 2 ,
a 3 ,...into the trial solution gives:y=x^1{
a 0 +a 0
( 1 × 3 )x+a 0
( 1 × 2 )×( 3 × 5 )x^2+a 0
( 1 × 2 × 3 )×( 3 × 5 × 7 )x^3+a 0
( 1 × 2 × 3 × 4 )×( 3 × 5 × 7 × 9 )x^4+ ···}i.e.y=a 0 x^1{
1 +x
( 1 × 3 )+x^2
( 1 × 2 )×( 3 × 5 )+x^3
( 1 × 2 × 3 )×( 3 × 5 × 7 )+x^4
( 1 × 2 × 3 × 4 )×( 3 × 5 × 7 × 9 )+ ···}
(26)(b) Withc=1
2ar=ar− 1
2 (c+r− 1 )(c+r)−(c+r)+ 1
from equation (25)i.e.ar=ar− 12(
1
2+r− 1)(
1
2+r)
−(
1
2+r)
+ 1=ar− 12(
r−1
2)(
r+1
2)
−1
2−r+ 1=ar− 12(
r^2 −1
4)
−1
2−r+ 1