Power series methods of solving ordinary differential equations 503
− a 0 x
c+1
− a 1 x
c+2
− a 2 x
c+3
− a 3 x
c+4
− · · · − a r x
c+r+1
− · · ·
(c)
(iv) The indicial equation, which is obtained by
equating the coefficient of the lowest power of
x to zero, gives the value(s) of c. Equating the
total coefficients of x c (from equations (a) to (c))
to zero gives:
2a 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 − 2c − c + 1] = 0
i.e.
a 0 [2c 2 − 3c + 1] = 0
i.e.
a 0 [(2c − 1)(c − 1)] = 0
from which,
c = 1 or c =
1
2
The coefficient of the general term, i.e. x c+r ,
gives (from equations (a) to (c)):
2a r (c + r − 1)(c + r) − a r (c + r)
+ a r − a r−1 = 0
from which,
a r [2(c + r − 1)(c + r) − (c + r) + 1] = a r−1
and a r =
a r−1
2(c +r −1)(c +r)−(c +r) +1
(25)
(a) With c = 1, a r =
a r−1
2(r)(1 + r) − (1 + r ) +1
=
a r−1
2r + 2r 2 − 1 − r + 1
=
a r−1
2r 2 + r
=
a r−1
r (2r + 1)
Thus, when r = 1,
a 1 =
a 0
1(2 + 1)
=
a 0
1 × 3
when r = 2,
a 2 =
a 1
2(4 + 1)
=
a 1
(2 × 5)
=
a 0
(1 × 3)(2 × 5)
or
a 0
(1 × 2) × (3 × 5)
when r = 3,
a 3 =
a 2
3(6 + 1)
=
a 2
3 × 7
=
a 0
(1 × 2 × 3) × (3 × 5 × 7)
when r = 4,
a 4 =
a 3
4(8 + 1)
=
a 3
4 × 9
=
a 0
(1 × 2 × 3 × 4) × (3 × 5 × 7 × 9)
and so on.
From equation (23), the trial solution was:
y = x
c
a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+ · · ·
+ a r x
r
+ · · ·
Substituting c = 1 and the above values of a 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) With c =
1
2
a r =
a r−1
2(c + r − 1)(c + r) − (c + r) + 1
from equation (25)
i.e. a r =
a r−1
2
1
2
+ r − 1
1
2
+ r
−
1
2
+ r
+1
=
a r−1
2
r −
1
2
r +
1
2
−
1
2
− r + 1
=
a r−1
2
r 2 −
1
4
−
1
2
− r + 1