Page 67 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 57
2
Equating x terms, we find that b 0 =−1. For higher order terms
3
= 2b 2 + b 1 = 2b 2 + b 1
2
Taking b 1 = 0,
3
b 2 =
4
5 3
− = 6b 3 + b 2 = 6b 3 +
12 4
7
b 3 =−
36
The second solution is
3 3 7 4

u 2 = u 1 ln(x) − x − x + x − ... (4.65)
4 36
The complete solution is therefore

3 7

3
4
u = [C 1 + C 2 ln x] u 1 − C 2 x − x + x − ··· (4.66)
4 36
Problems
1. Find two Frobenius series solutions
2

2
x u + 2xu + (x − 2)u = 0
2. Find two Frobenious series solutions
1
2 2 u = 0
x u + xu + x −
4
3. Show that the indicial equation for the differential equation

xu + u + xu = 0

has roots s =−1 and that the differential equation has only one Frobenius series
solution. Find that solution. Then find another solution in the form
∞ ∞
n+s s +m
u = ln c n x + a n x
n=0 m=0

where the first summation above is the first Frobenius solution.