Page 151 - Advanced engineering mathematics

4.2 Frobenius Solutions 131

Some of these coefficients are
3 6
c 1 =− c 0 =− c 0 ,
5/2 5
5 5 6 6
c 2 =− c 1 =− − c 0 = c 0 ,
7 7 5 7
7 14 6 4
c 3 =− c 2 =− c 0 =− c 0 ,
27/2 27 7 9

and so on. One Frobenius solution is

6 2 6 3 4 4
y 1 (x) = c 0 x − x + x − x + ··· .
5 7 9
Because r 1 is a nonnegative integer, this first Frobenius series is actually a power series about 0.
For a second Frobenius solution, substitute r = r 2 =−1/2 into the recurrence relation. To
∗
avoid confusion with the first solution, we will denote the coefficients c instead of c n . We obtain
n
3
1 + 2 n −
∗ 2 ∗
n 1 3 1 1 1 n−1
c =−

c
n − n − + n − −
2 2 2 2 2
for n = 1,2,3,···. This simplifies to
2n − 2
∗
c ∗
n 3 n−1
c =−
n n −
2
∗
for n = 1,2,3,···. It happens in this example that c = 0, so each c = 0for n = 1,2,3,···, and
∗
1
n
the second Frobenius solution is
∞

∗
∗
y 2 (x) = c x n−1/2 = c x −1/2
n 0
n=0
for x > 0.
EXAMPLE 4.6 Case 3 of the Frobenius Theorem
We will solve
2
x y + 5xy + (x + 4)y = 0.

In Example 4.5, we found the indicial equation
r(r − 1) + 5r + 4 = 0
with repeated root r 1 =r 2 =−2 and the recurrence relation
1
c n =− c n−1
n 2
for n = 1,2,···. This yielded the first Frobenius solution
∞
n 1 n−2
(−1) x
y 1 (x) = c 0
(n!) 2
n=0

1 1 1 2
−2
−1
= c 0 x − x + − x + x + ··· .
4 36 576
Conclusion (3) of Theorem 4.2 tells us the general form of a second solution that is linearly
independent from y 1 (x). Set
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October 14, 2010 14:17 THM/NEIL Page-131 27410_04_ch04_p121-136