Page 155 - Advanced engineering mathematics

4.2 Frobenius Solutions 135

Furthermore, the recurrence relation is
1 (2n + 1)k
∗
c ∗ = c −
n+1 n
n(n + 1) n!(n + 1)!
for n = 1,2,···. Since c can be any nonzero number, we will for convenience let c = 1. For a
∗
∗
0 0
particular solution, we may also choose c = 1. These give us
∗
1
3 2 7 3 35 4
y 2 (x) = y 1 ln(x) + 1 − x − x − x − ··· .
4 36 1728
SECTION 4.2 PROBLEMS

In each of Problems 1 through 10, find the first five terms 5. 4xy + 2y + 2y = 0

of each of two linearly independent solutions. 2
6. 4x y + 4xy − y = 0

1. xy + (1 − x)y + y = 0 7. x y − 2xy − (x − 2)y = 0
2
2

2. xy − 2xy + 2y = 0 8. xy − y + 2y = 0

3. x(x − 1)y + 3y − 2y = 0 9. x(2 − x)y − 2(x − 1)y + 2y = 0

2
2
4. 4x y + 4xy + (4x − 9)y = 0 10. x y + x(x + 1)y − y = 0

2
3

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October 14, 2010 14:17 THM/NEIL Page-135 27410_04_ch04_p121-136