Page 63 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 53
Thefirsttermisthe indicial equation.

r(r − 1) − 2r + 2 = 0 (4.43)

There are two distinct roots, r 1 = 2and r 2 = 1. However they differ by an integer.

r 1 − r 2 = 1.

Substituting r 1 = 2 into (4.39) and noting that each coefficient of x n+r must be zero,

[(n + 2)(n + 1) − 2(n + 2) + 2]c n + c n−1 = 0 (4.44)

The recurrence equation is
−c n−1
c n =
(n + 2)(n − 1) + 2
−c 0
c 1 =
2
−c 1 c 0
c 2 = = c 0
6 12
−c 2 −c 0
c 3 = = (4.45)
12 144
The first Frobenius series is therefore
1 1 1
2 3 4 5
u 1 = c 0 x − x + x − x + ... (4.46)
2 12 144
We now attempt to find the Frobenius series corresponding to r 2 = 1. Substituting into (4.44)
we find that

[n(n + 1) − 2(n + 1) + 2]c n =−c n−1 (4.47)
When n = 1, c 0 must be zero. Hence c n must be zero for all n and the attempt to find a second
Frobenius series has failed. This will not always be the case when roots differ by an integer as
illustrated in the following example.

Example 4.6 (roots differing by an integer 2). Consider the differential equation

2
x u + x u − 2u = 0 (4.48)
2
2
You may show that the indicial equation is r − r − 2 = 0 with roots r 1 = 2, r 2 =−1and the
roots differ by an integer. When r = 2 the recurrence equation is
n + 1
c n =− c n−1 (4.49)
n(n + 3)