Page 152 - Advanced engineering mathematics

132 CHAPTER 4 Series Solutions

∞
n−2
∗
y 2 (x) = y 1 (x)ln(x) + c x .
n
n=1
Note that on the right, the series starts at n =1, not n =0. Substitute this series into the differential
equation and find after some rearranging of terms that
∞ ∞

∗ n−2 ∗ n−2
4y 1 + 2xy + (n − 2)(n − 3)c x + 5(n − 2)c x
n
1
n
n=1 n=1
∞ ∞
n−1 n−2
∗
∗
+ c x + 4c x
n
n
n=1 n=1
2

+ ln(x) x y + 5xy + (x + 4)y 1 = 0.
1
1
∗
The bracketed coefficient of ln(x) is zero because y 1 is a solution. Choose c = 1 (we need only
0
∞ n−1 ∞ n−2
one second solution), shift the indices to write n=1 c x = n=2 c ∗ n−1 x , and substitute the
∗
n
series for y 1 (x) to obtain
∞ n n
4(−1) 2(−1)
∗
− 2x −1 + c x −1 + + (n − 2) + (n − 2)(n − 3)c ∗
1 (n!) 2 (n!) 2 n
n=2
n−2
+5(n − 2)c + c ∗ + 4c x = 0.
∗
∗
n n−1 n
−1
Set the coefficient of each power of x equal to 0. From the coefficient of x ,wehave c = 2.
∗
1
From the coefficient of x n−2 , we obtain (after some routine algebra)
2(−1) n
2 ∗
n + n c + c ∗ = 0
(n!) 2 n n−1
or
1 2(−1) n
∗ ∗
c =− c −
n 2 n−1 2
n n(n!)
for n = 2,3,4,···. With this, we can calculate as many coefficients as we want, yielding
2 3 11
y 2 (x) = y 1 (x)ln(x) + − + x
x 4 108
25 2 137 3
− x + x + ··· .
3456 432,000
The next two examples illustrate Case (4) of the theorem, first with k = 0 and then k = 0.
EXAMPLE 4.7 Case 4 of Theorem 4.2 with k = 0
We will solve
2
2
x y + x y − 2y = 0.

∞ n+r

There is a regular singular point at 0. Substitute y = n=0 c n x to obtain
(r(r − 1) − 2)c 0 x r
∞
n+r
+ [(n +r)(n +r − 1)c n + (n +r − 1)c n−1 − 2c n ]x = 0.
n=1
2
The indicial equation is r −r − 2=0 with roots r 1 =2,r 2 =−1. Now r 1 −r 2 =3, putting us
in Case (4) of the theorem. From the coefficient of x n+r , we obtain the general recurrence relation
(n +r)(n +r − 1)c n + (+r − 1)c n−1 − 2c n = 0
for n = 1,2,3,···.
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October 14, 2010 14:17 THM/NEIL Page-132 27410_04_ch04_p121-136