Page 64 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

54 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
The first Frobenius series is

1 3 1

2
3
u 1 = c 0 x 2 1 − x + x − x + ... (4.50)
2 20 30
When r =−1 the recurrence equation is
[(n − 1)(n − 2) − 2]b n + (n − 2)b n−1 = 0 (4.51)

When n = 3 this results in b 2 = 0. Thus b n = 0 for all n ≥ 2 and the second series terminates.
1 1

u 2 = b 0 − (4.52)
x 2

4.1.5 Lessons: Finding Series Solution for Differential Equations with Regular
Singular Points
1. Assume a solution of the form

∞
n+r
u = c n x , c 0 = 0 (4.53)
n=0
Differentiate term by term and insert into the differential equation. Set the coefficient
of the lowest power of x to zero to obtain a quadratic equation on r.
If the indicial equation yields two roots that do not differ by an integer there will
always be two Frobenius series, one for each root of the indicial equation.
2. If the roots are the same (repeated roots) the form of the second solution will be

∞

u 2 = u 1 ln(x) + b n x n+r 1 (4.54)
n=1
This equation is substituted into the differential equation to determine b n .
3. If the roots differ by an integer, choose the largest root to obtain a Frobenius series for
u 1 . The second solution may be another Frobenius series. If the method fails assume a
solution of the form

∞

u 2 = u 1 ln(x) + b n x n+r 2 (4.55)
n=1
This equation is substituted into the differential equation to find b n .

This is considered in the next section.