Page 142 - Advanced engineering mathematics

122 CHAPTER 4 Series Solutions

Here n! (n factorial) is the product of the integers from 1 through n if n is a positive integer,
and 0!= 1 by definition. The symbol f (n) (x 0 ) denotes the nth derivative of f evaluated at x 0 .As
examples of power series representations, sin(x) expanded about 0 is
∞
1
2n+1
sin(x) = x
(2n + 1)!
n=0
for all x, and the geometric series is
∞
1 n
= x
1 − x
n=0
for −1 < x < 1.
An initial value problem having analytic coefficients has analytic solutions. We will state
this for the first- and second-order cases when the differential equation is linear.

THEOREM 4.1
1. If p and q are analytic at x 0 , then the problem

y + p(x)y = q(x); y(x 0 ) = y 0
has a unique solution that is analytic at x 0 .
2. If p, q, and f are analytic at x 0 , then the problem

y + p(x)y + q(x)y = f (x); y(x 0 ) = A, y (x 0 ) = B
has a unique solution that is analytic at x 0 .
We are therefore justified in seeking power series solutions of linear equations having ana-

∞ n
lytic coefficients. This strategy may be carried out by substituting y = n=0 a n (x − x 0 ) into the
differential equation and attempting to solve for the a n s.

EXAMPLE 4.1
We will solve
1

y + 2xy = .
1 − x
We can solve this using an integrating factor, obtaining
x 1 2 2
2
y(x) = e −x e −ξ dξ + ce −x .
0 1 − ξ
This is correct, but it involves an integral we cannot evaluate in closed form. For a series
solution, let
∞

n
y = a n x .
n=0
Then
∞

n−1
y = na n x
n=1
with the summation starting at 1, because the derivative of the first term a 0 of the
power series for y is zero. Substitute the series into the differential equation to obtain

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October 14, 2010 14:17 THM/NEIL Page-122 27410_04_ch04_p121-136