Page 144 - Advanced engineering mathematics

124 CHAPTER 4 Series Solutions

1
(n = 3) a 4 = (1 − 2a 2 )
4
1 1
= (1 − 1 + 2a 0 ) = a 0 ,
4 2
1 1 1
(n = 4) a 5 = (1 − 2a 3 ) = (1 + 2/3) = ,
5 5 3
1 1 − a 0
(n = 5) a 6 = (1 − 2a 4 ) = ,
6 6
1 1
(n = 6) a 7 = (1 − 2a 5 ) = ,
7 21
and so on. With the coefficients computed thus far, the solution has the form
1 1
2
y(x) = a 0 + x + (1 − 2a 0 )x − x 3
2 3
1 4 1 5
+ a 0 x + x
2 3
1 6 1 7
+ (1 − a 0 )x + x + ··· .
6 21
This has one arbitrary constant, a 0 , as expected. By continuing to use the recurrence relation, we
can compute as many terms of the series as we like.

EXAMPLE 4.2

We will find a power series solution of
2
y + x y = 0

expanded about x 0 = 0.

∞ n
Substitute y = a n x into the differential equation. This will require that we compute
n=0
∞ ∞

n−1 n−2
y = na n x and y = (n − 1)na n x .
n=1 n=2
Substitute these power series into the differential equation to obtain
∞ ∞
n−2 2 n
(n − 1)na n x + x a n x = 0
n=2 n=0
or
∞ ∞
n−2 n+2
n(n − 1)a n x + a n x = 0. (4.5)
n=2 n=0
We will shift indices so that the power of x in both summations is the same, allowing us to
combine terms from both summations. One way to do this is to write
∞ ∞
n−2 n
(n − 1)na n x = (n + 2)(n + 1)a n+2 x
n=2 n=0
and
∞ ∞
n+2 n
a n x = a n−2 x .
n=0 n=2

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October 14, 2010 14:17 THM/NEIL Page-124 27410_04_ch04_p121-136