I would imagine no one has replied to this problem because power series solutions of DE's are tedious.
But, here is a little help if you still need it.
Use \(\displaystyle y=\sum_{n=0}^{\infty}c_{n}x^{k}\) and find the first and second derivatives.
Then, sub these into the given DE.
This gives:
\(\displaystyle \displaystyle \underbrace{\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}}_{\text{k=n-2}}+\underbrace{\sum_{n=1}^{\infty}nc_{n}x^{n}}_{\text{k=n}}+\underbrace{\sum_{n=1}^{\infty}nc_{n}x^{n-1}}_{\text{k=n-1}}-\underbrace{2\sum_{n=0}^{\infty}c_{n}x^{n}}_{\text{k=n}}=0\)
This shifts the index to:
\(\displaystyle \displaystyle \sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2}x^{k}+\)\(\displaystyle \displaystyle \sum_{k=1}^{\infty}kc_{k}x^{k}\)\(\displaystyle \displaystyle +\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}-2\sum_{k=0}^{\infty}c_{k}x^{k}\)
We now want all of the indices to begin at k=1. So, we take out the k=0 case where applicable.
\(\displaystyle \displaystyle 2c_{2}+c_{1}-2c_{0}+\sum_{k=1}^{\infty}\left[(k+2)(k+1)c_{k+2}+(k+1)c_{k+1}+(k-2)c_{k}\right]x^{k}\)
Now, try continuing with the recurrence to solve the DE.
