Page 62 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

52 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and the roots of this equation are both r = 1. Setting the last two sums to zero we find the
recurrence equation

1
c n =− c n−1 (4.37)
(n + r − 1)(n + r) − (n + r) + 1
and since r = 1,

1
c n =− c n−1 (4.38)
n(n + 1) − (n + 1) + 1
c 1 =−c 0
−1 1
c 2 = c 1 = c 0
6 − 3 + 1 4
−1 −1 −1
c 3 = c 2 = c 1 = c 0
12 − 4 + 1 9 36
etc.
The Frobenius series is

1 1

2 3 4
u 1 = c 0 x − x + x − x + ... (4.39)
4 36
In this case there is no second solution in the form of a Frobenius series because of the repeated
root. We shall soon see what form the second solution takes.

Example 4.5 (roots differing by an integer 1). Next consider the equation

x u − 2xu + (x + 2)u = 0 (4.40)

2
There is a regular singular point at x = 0. We therefore expect a solution in the form of the
Frobenius series (4.18). Substituting (4.18), (4.19), (4.20) into our differential equation, we
obtain
∞ ∞ ∞ ∞
n+r n+r n+r n+r+1
(n + r)(n + r − 1)c n x − 2(n + r)c n x + 2c n x + c n x = 0
n=0 n=0 n=0 n=0
(4.41)
Taking out the n = 0 term and shifting the last summation,

∞
r n+r
[r(r − 1) − 2r + 2]c 0 x + [(n + r)(n + r − 1) − 2(n + r) + 2]c n x
n=1
∞
n+r
+ c n−1 x = 0 (4.42)
n=1