So this is the problem:
xy^{''} + y^{'}- xy = 0
Find the first nonzero terms in a Frobenius series solution of the given differential equation. Then use the reduction of order technique to find the logarithmic term and the first three nonzero terms in a second linearly independent solutions.
I did the first part:
n(n+1)c_{n+1}x^{n} + (n+1)c_{n+1}x^{n} - c_{n-1}x^{n}= 0
[n(n+1) + (n+1)]c_{n+1} = c_{n-1}
c_{n+1} = c_{n-1}/ n(n+1) + (n+1)
c_{1} = 0
c_{2} = c_{0}/2^{2 }c_{3} = c_{1}/3^{2 }= 0
c4 = c_{2}/2^{2}*4^{2 c}_{5}= 0
y_{1}(x) = 1 + x^{2}/2^{2} + x^{4}/2^{2}*4^{2} + x^{6}/2^{2}*4^{2}*6^{2} + x^{8}/2^{2}*4^{2}*6^{2}*8^{2} + ...
OK, and now to find y_{2}, I'm kind of stuck. Since in the indicial equation, the roots are both r=0, so the equation I should use is: y_{2} = y_{1}*ln(x) + x^{r1+1} b_{n}x^{n}.
The solution manual of my text book starts off with this:
y_{2} = y_{1} x^{-1} * (1 + x^{2}/2^{2} + x^{4}/2^{2}*4^{2} + x^{6}/2^{2}*4^{2}*6^{2} + x^{8}/2^{2}*4^{2}*6^{2}*8^{2})^{-2 }dx
... I have NO clue where that came from. Someone please explain how to solve this? I'm studying for an exam and an explanation would be immensely helpful.