Power series methods of solving ordinary differential equations 507
i.e. y=a 0 xc+a 1 xc+^1 +a 2 xc+^2 +a 3 xc+^3
+ ···+arxc+r+··· (35)(ii) Differentiating equation (35) gives:y′=a 0 cxc−^1 +a 1 (c+ 1 )xc
+a 2 (c+ 2 )xc+^1 +···
+ar(c+r)xc+r−^1 +···
andy′′=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1
+a 2 (c+ 1 )(c+ 2 )xc+···
+ar(c+r− 1 )(c+r)xc+r−^2 +···(iii) Substitutingy, y′ and y′′ into each term of
the given equation:x^2 y′′+xy′+(x^2 −v^2 )y= 0
gives:
a 0 c(c− 1 )xc+a 1 c(c+ 1 )xc+^1
+a 2 (c+ 1 )(c+ 2 )xc+^2 +···
+ar(c+r− 1 )(c+r)xc+r+···+a 0 cxc
+a 1 (c+ 1 )xc+^1 +a 2 (c+ 2 )xc+^2 +···
+ar(c+r)xc+r+···+a 0 xc+^2 +a 1 xc+^3
+a 2 xc+^4 +···+arxc+r+^2 +···−a 0 v^2 xc
−a 1 v^2 xc+^1 −···−arv^2 xc+r+···= 0
(36)
(iv) Theindicial equationis obtained by equating
the coefficient of the lowest power ofxto zero.
Hence, a 0 c(c− 1 )+a 0 c−a 0 v^2 = 0
from which, a 0 [c^2 −c+c−v^2 ]= 0
i.e. a 0 [c^2 −v^2 ]= 0
from which, c=+vorc=−vsincea 0 = 0
For the term inxc+r,
ar(c+r− 1 )(c+r)+ar(c+r)+ar− 2
−arv^2 = 0
ar[(c+r− 1 )(c+r)+(c+r)−v^2 ]=−ar− 2
i.e. ar[(c+r)(c+r− 1 + 1 )−v^2 ]=−ar− 2
i.e. ar[(c+r)^2 −v^2 ]=−ar− 2
i.e. therecurrence relationis:ar=ar− 2
v^2 −(c+r)^2for r≥ 2 (37)For the term inxc+^1 ,
a 1 [c(c+ 1 )+(c+ 1 )−v^2 ]= 0
i.e. a 1 [(c+ 1 )^2 −v^2 ]= 0
but ifc=v a 1 [(v+ 1 )^2 −v^2 ]= 0i.e. a 1 [2v+1]= 0
Similarly, ifc=−va 1 [1− 2 v]= 0
The terms( 2 v+ 1 )and( 1 − 2 v)cannot both be
zero sincevis a real constant, hencea 1 =0.
Since a 1 =0, then from equation (37)
a 3 =a 5 =a 7 =...= 0
anda 2 =
a 0
v^2 −(c+ 2 )^2a 4 =a 0
[v^2 −(c+ 2 )^2 ][v^2 −(c+ 4 )^2 ]a 6 =a 0
[v^2 −(c+ 2 )^2 ][v^2 −(c+ 4 )^2 ][v^2 −(c+ 6 )^2 ]
and so on.Whenc=+v,a 2 =a 0
v^2 −(v+ 2 )^2=a 0
v^2 −v^2 − 4 v− 4=−a 0
4 + 4 v=−a 0
22 (v+ 1 )a 4 =a 0
[
v^2 −(v+ 2 )^2][
v^2 −(v+ 4 )^2]=a 0
[− 22 (v+ 1 )][− 23 (v+ 2 )]=a 0
25 (v+ 1 )(v+ 2 )=a 0
24 × 2 (v+ 1 )(v+ 2 )a 6 =a 0
[v^2 −(v+ 2 )^2 ][v^2 −(v+ 4 )^2 ][v^2 −(v+ 6 )^2 ]=a 0
[2^4 × 2 (v+ 1 )(v+ 2 )][− 12 (v+ 3 )]=−a 0
24 × 2 (v+ 1 )(v+ 2 )× 22 × 3 (v+ 3 )=−a 0
26 ×3!(v+ 1 )(v+ 2 )(v+ 3 )and so on.The resulting solution forc=+vis given by:y=u=Axv{
1 −x^2
22 (v+ 1 )+x^4
24 ×2!(v+ 1 )(v+ 2 )−x^6
26 ×3!(v+ 1 )(v+ 2 )(v+ 3 )+···}(38)