Page 150 - Advanced engineering mathematics

130 CHAPTER 4 Series Solutions

EXAMPLE 4.5 Case 2 of the Frobenius Theorem
We will solve
1 1

2

x y + x + 2x y + x − y = 0.
2 2
It is routine to check that 0 is a regular singular point. Substitute the Frobenius series y =
∞ c n x n+r to obtain

n=0
∞ ∞ ∞
n+r−2 1 n+r n+r+1
(n +r)(n +r − 1)c n c + (n +r)c n x + 2(n +r)x
2
n=0 n=0 n=0
∞ ∞
n+r+1 1 n+r
+ c n x − c n x = 0.
2
n=0 n=0
In order to be able to factor x n+r from most terms, shift indices in the third and fourth summations
to write this equation as
∞
1 1 n+r
(n +r)(n +r − 1)c n + (n +r)c n + 2(n +r − 1)c n−1 + c n−1 − c n x
2 2
n=1
1 1

r
+ r(r − 1)c 0 + c 0 r − c 0 x = 0.
2 2
This equation will hold if the coefficient of each power of x is zero:
1 1

r(r − 1) + r − c 0 = 0 (4.9)
2 2
and for n = 1,2,3,···,
1 1
(n +r)(n +r − 1)c n + (n +r)c n + 2(n +r − 1)c n−1 + c n−1 − c n = 0. (4.10)
2 2
Assuming that c 0 = 0, an essential requirement of the method, equation (4.9) implies that
1 1
r(r − 1) + r − = 0. (4.11)
2 2
This is the indicial equation for this differential equation. It has the roots r 1 = 1 and r 2 =−1/2.
This puts us in case 2 of the Frobenius theorem. From equation (4.10), we obtain the recurrence
relation
1 + 2(n +r − 1)
c n =− c n−1
1
(n +r)(n +r − 1) + (n +r) − 1
2 2
for n = 1,2,3,···.
First put r 1 = 1 into the recurrence relation to obtain
2n + 1

c n−1
c n =− 3
n n +
2
for n = 1,2,3,···.

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October 14, 2010 14:17 THM/NEIL Page-130 27410_04_ch04_p121-136