Page 61 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 51
Computing a few of the coefficients,

3 6
c 1 =− c 0 =− c 0
5 5
2
5 6
c 2 =− c 1 =− c 0
7 7
7 4
c 3 =− c 2 =− c 0
27 9
2
etc. and the first Frobenius series is
6 2 6 3 4 4

u 1 = c 0 x − x + x − x + ··· (4.30)
5 7 9
Setting r =−1/2 in the recurrence equation (4.26) and using b n instead of c n to distinguish it
from the first case,
2n − 2
b n−1 (4.31)
3
b n =−
n n −
2
Noting that in this case b 1 = 0, all the following b n s must be zero and the second Frobenius
series has only one term: b 0 x −1/2 . The complete solution is
6 2 6 3 4 4 −1/2

u = c 0 x − x + x − x + ··· + b 0 x (4.32)
5 7 9
Example 4.4 (repeated roots). Next consider the differential equation

x u − xu + (x + 1)u = 0 (4.33)
2
There is a regular singular point at x = 0, so we attempt a Frobenius series around x = 0.

Differentiating (4.17) and substituting into (4.30),
∞ ∞ ∞ ∞
n+r n+r n+r n+r+1
(n + r − 1)(n + r)c n x − (n + r)c n x + c n x + c n x = 0 (4.34)
n=0 n=0 n=0 n=0
or
∞ ∞

r n+r n+r
[r(r − 1) − r + 1]c 0 x + [(n + r − 1)(n + r) − (n + r) + 1]c n x + c n−1 x = 0
n=1 n=1
(4.35)
where we have shifted the index in the last sum.
The indicial equation is

r(r − 1) − r + 1 = 0 (4.36)