Page 65 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 55
4.1.6 Logarithms and Second Solutions
Example 4.7. Reconsider Example 4.4 and assume a solution in the form of (4.54). Recall
that in Example 4.4 the differential equation was

x u − xu + (1 + x)u = 0 (4.56)

2
and the indicial equation yielded a double root at r = 1.
A single Frobenius series was

x 3 x 4
2
u 1 = x − x + − + ···
4 36
Now differentiate Eq. (4.54).

∞
1
n+r−1
u = u ln x + u 1 + (n + r)b n x
2
1
x
n=1
∞
2 1
n+r−2
u = u ln x + u − u 1 + (n + r − 1)(n + r)b n x (4.57)
2 1 1 2
x x
n=1
Inserting this into the differential equation gives
ln(x)[x u − xu + (1 + x)u 1 ] + 2(xu − u 1 )
2

1
1
1
∞
n+r n+r n+r
+ [b n (n + r − 1)(n + r)x − b n (n + r)x + b n x ]
n=1
∞
n+r+1
+ b n x = 0 (4.58)
n=1
The first term on the left-hand side of (4.52) is clearly zero because the term in brackets is the
original equation. Noting that r = 1 in this case and substituting from the Frobenius series for
u 1 ,wefind (c 0 can be set equal to unity without losing generality)
x 3 x 4 ∞ ∞
2 n+1 n+1
2 −x + − + ··· + [n(n + 1) − (n + 1) + 1]b n x + b n−1 x = 0
3 12
n=1 n=2
(4.59)
or
x 4 ∞
2 3 2 2
n+1
−2x + x − + ··· + b 1 x + n b n + b n−1 x = 0 (4.60)
6
n=2
Equating coefficients of x raised to powers we find that b 1 = 2