Page 66 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

56 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
For n ≥ 2

1 + 4b 2 + b 1 = 0 b 2 =−3/4
1 11
− + 9b 3 + b 2 = 0 b 3 =
6 108
etc.
3 11

2 3 4
u 2 = u 1 ln x + 2x − x + x − ··· (4.61)
4 108
The complete solution is
3 11
2 3 4
u = [C 1 + C 2 ln x] u 1 + C 2 2x − x + x − ··· (4.62)
4 108
Example 4.8. Reconsider Example 4.5 in which a second Frobenius series could not be found
because the roots of the indicial equation differed by an integer. We attempt a second solution
in the form of (4.55).
The differential equation in Example 4.5 was

2

x u − 2xu + (x + 2)u = 0
and the roots of the indicial equation were r = 2and r = 1, and are therefore separated by an
integer. We found one Frobenius series
1 1 1
2 3 4 5
u 1 = x − x + x − x + ···
2 12 144
for the root r = 2, but were unable to find another Frobenius series for the case of r = 1.
Assume a second solution of the form in Eq. (4.55). Differentiating and substituting into
(4.40)

2
[x u − 2xu + (x + 2)u] ln(x) + 2xu − 3u 1
1
∞
n+r
+ b n [(n + r)(n + r − 1) − 2(n + r) + 2]x
n=1
∞
n+r+1
+ b n x = 0 (4.63)
n=1
Noting that the first term in the brackets is zero, inserting u 1 and u from (4.50) and noting

1
that r 2 = 1
∞
3 5 7
2 3 4 5 2 n+1
x − x + x − x + ... + b 0 x + {[n(n − 1)]b n + b n−1 }x = 0 (4.64)
2 12 144
n=2