Page 153 - Advanced engineering mathematics

4.2 Frobenius Solutions 133

For a first solution, use r = 2 to obtain the recurrence relation
n + 1
c n =− c n−1
n(n + 3)
for n = 1,2,···. Using this, we obtain a first solution
1 3 1 1 1

4
3
5
2
y 1 (x) = c 0 x 2 1 − x + x − x + x − x + ··· .
2 20 30 168 1120
Now we need a second, linearly independent solution. Put r =−1 into the general recurrence
relation to obtain
∗
(n − 1)(n − 2)c + (n − 2)c ∗ − 2c = 0
∗
n n−1 n
for n = 1,2,···. When n = 3, this gives c = 0, which forces c = 0for n = 2,3,···. But then
∗
∗
2 n
1
y 2 (x) = c ∗ + c .
∗
0 1
x
Substitute this into the differential equation to obtain

1
2
−2
2
−3
∗
∗
x (2c x ) + x (−c x ) − 2 c + c ∗ 0 =−c − 2c = 0.
∗
∗
∗
1
0
0
0
1
x
Then c =−c /2, and a second solution is
∗
∗
1 0
1 1

y 2 (x) = c ∗ −
0
x 2
with c arbitrary but nonzero. The functions y 1 and y 2 form a fundamental set of solutions. In
∗
0
these solutions, there is no y 1 (x)ln(x) term.
EXAMPLE 4.8 Case 4 of Theorem 4.2 with k = 0
We will solve

xy − y = 0,

∞ n+r
which has a regular singular point at 0. Substitute y = c n x and rearrange terms to obtain
n=0
∞
n+r−1
2
r−1
(r −r)c 0 x + [(n +r)(n +r − 1)c n − c n−1 ]x = 0.
n=1
The indicial equation is r − r = 0, with roots r 1 = 1,r 2 = 0. Here r 1 − r 2 = 1, a positive integer,
2
putting us in Case (4) of the theorem. The general recurrence relation is
(n +r)(n +r − 1)c n − c n−1 = 0
for n = 1,2,···. With r = 1, this is
1
c n = c n−1 ,
n(n + 1)
and some of the coefficients are
1
c 1 = c 0 ,
2
1 1
c 2 = c 1 = c 0 ,
2(3) 2(2)(3)
1 1
c 3 = c 2 = c 0 ,
3(4) 2(3)(2)(3)(4)
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October 14, 2010 14:17 THM/NEIL Page-133 27410_04_ch04_p121-136