Ok. I understand and could get to common out the x terms, but am stuck at this step.
(4n-5)/2\sum a_{n-1}x^{(4n+1)/4}+4n\sum a_{n}x^{(4n+1)/4} = x^{2}+2x
I was looking up the answers to the values for coefficients and they come out to be(for non-homogeneous part)
a_{0} = 0,
a_{1} = 2/3...
Thanks for the suggestions/comments.
@coldatoms - to get the indicial equation, how do we get rid of the right hand side term in equation 1.
i am still left out with a x or an x^2 when i use the frobenius method.
I am having trouble getting to a solution for this differential equation
2(x^2+2x)y' - y(x+1) = x^2+2x -------- 1
for a series solution, we have to assume y = \sum a_{n}x^n ---------- 2
if we divide equation 1 by x^2 + 2x , we get (x+1)/(x^2+2x) for the y term, which is where my problem...