Power series methods of solving ordinary differential equations 507
i.e. y = a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+· · ·
(35)
(ii) Differentiating equation (35) gives:
y
= a 0 cx
c−1
+ a 1 (c + 1)x
c
+ a 2 (c + 2)x
c+1
+· · ·
+ a r (c + r)x
c+r−1
+ · · ·
and y
= a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+· · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
(iii) Substituting y, y and y into each term of
the given equation: x 2 y + x y + (x 2 − v 2 )y = 0
gives:
a 0 c(c − 1)x
c
+ a 1 c(c + 1)x
c+1
+ a 2 (c + 1)(c + 2)x
c+2
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r
+ · · · + a 0 cx
c
+ a 1 (c + 1)x
c+1
+ a 2 (c + 2)x
c+2
+ · · ·
+ a r (c + r)x
c+r
+ · · · + a 0 x
c+2
+ a 1 x
c+3
+ a 2 x
c+4
+ · · · + a r x
c+r+2
+ · · · − a 0 v
2 x
c
− a 1 v
2 x
c+1
− · · · − a r v
2 x
c+r
+ · · · = 0
(36)
(iv) The indicial equation is obtained by equating
the coefficient of the lowest power of x to 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 =+v or c =−v since a 0 = 0
For the term in x c+r ,
a r (c + r − 1)(c + r) + a r (c + r) + a r−2
− a r v
2
= 0
a r [(c + r − 1)(c + r) + (c + r) − v
2 ] =−a r−2
i.e. a r [(c + r)(c + r − 1 + 1) − v 2 ] =−a r−2
i.e.
a r [(c + r) 2 − v 2 ] =−a r−2
i.e. the recurrence relation is:
a r =
a r−2
v 2 − (c + r)
2
for r ≥ 2
(37)
For the term in x c+1 ,
a 1 [c(c + 1) + (c + 1) − v
2 ] = 0
i.e.
a 1 [(c + 1) 2 − v 2 ] = 0
but if c = v
a 1 [(v + 1) 2 − v 2 ] = 0
i.e.
a 1 [2v + 1] = 0
Similarly, if c = −va 1 [1 − 2v] = 0
The terms (2v + 1) and (1 − 2v) cannot both be
zero since v is a real constant, hence a 1 = 0.
Since a 1 = 0, then from equation (37)
a 3 = a 5 = a 7 = ... = 0
and
a 2 =
a 0
v 2 − (c + 2) 2
a 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.
When c =+v,
a 2 =
a 0
v 2 − (v + 2) 2 =
a 0
v 2 − v 2 − 4v − 4
=
−a 0
4 + 4v
=
−a 0
2 2 (v + 1)
a 4 =
a 0
v 2 − (v + 2) 2
v 2 − (v + 4) 2
=
a 0
[−2 2 (v + 1)][−2 3 (v + 2)]
=
a 0
2 5 (v + 1)(v + 2)
=
a 0
2 4 × 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
2 4 × 2(v + 1)(v + 2) × 2 2 × 3(v + 3)
=
−a 0
2 6 × 3! (v + 1)(v + 2)(v + 3)
and so on.
The resulting solution for c = +v is given by:
y = u =
A x
v
1−
x 2
2 2 (v +1)
+
x 4
2 4 × 2! (v +1)(v +2)
−
x 6
2 6 × 3! (v +1)(v + 2)(v + 3)
+· · ·
(38)