Page 148 - Advanced engineering mathematics

128 CHAPTER 4 Series Solutions

or, with r =−2,
(n − 2)(n − 3)c n + 5(n − 2)c n + c n−1 + 4c n = 0.
From this we obtain the recurrence relation
1
c n =− c n−1 for n = 1,2,··· .
(n − 2)(n − 3) + 5(n − 2) + 4
This simplifies to
1
c n =− c n−1 for n = 1,2,··· .
n 2
Solve for some coefficients:
c 1 =−c 0
1 1 1
c 2 =− c 1 = c 0 = c 0
4 4 2 2
1 1
c 3 =− c 2 =− c 0
9 (2 · 3) 2
1 1
c 4 =− c 3 = c 0
16 (2 · 3 · 4) 2
and so on. In general,
1
n
c n = (−1) c 0
(n!) 2
for n = 1,2,3,···. We have found the Frobenius solution
1 1 1 2

−1
−2
y(x) = c 0 x − x + − x + x + ···
4 36 576
∞
n 1 n−2
(−1) x
= c 0
(n!) 2
n=0
for x = 0. This series converges for all nonzero x.
Usually, we cannot expect the recurrence equation for c n to have such a simple form.
Example 4.4 shows that an equation with a regular singular point may have only one
Frobenius series solution about that point. A second, linearly independent solution is needed. The
following theorem tells us how to produce two linearly independent solutions. For convenience,
the statement is posed in terms of x 0 = 0.

THEOREM 4.2

Suppose 0 is a regular singular point of

P(x)y + Q(x)y + R(x)y = 0.
Then
(1) The differential equation has a Frobenius solution
∞
n+r
y(x) = c n x
n=0
with c 0 = 0. This series converges in some interval (0,h) or (−h,0).

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October 14, 2010 14:17 THM/NEIL Page-128 27410_04_ch04_p121-136