If \(\displaystyle r_{1}\) and \(\displaystyle r_{2}\) are distinct and do not differ by an integer, then there are two linearly independent solutions of the form:
\(\displaystyle y_{1}=\sum_{n=0}^{\infty}c_{n}x^{n+r_{1}}, \;\ c_{0}\neq 0\)
\(\displaystyle y_{2}=\sum_{n=0}^{\infty}b_{n}x^{n+r_{2}}, \;\ b_{0}\neq 0\)
Solving DE's using series is rather tedious to type out here.
But, we start with:
\(\displaystyle y=\sum_{n=0}^{\infty}c_{n}x^{n+r}\)
\(\displaystyle y'=\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}\)
\(\displaystyle y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r-2}\)
make the subs into the given DE:
\(\displaystyle x^{2}(x-1)\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r-2}-x\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}+(x+1)\sum_{n=0}^{\infty}c_{n}x^{n+r}=0\)
\(\displaystyle \sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r+1}-\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r}-\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r}+\sum_{n=0}^{\infty}c_{n}x^{n+r+1}+\sum_{n=0}^{\infty}c_{n}x^{n+r}=0\)
\(\displaystyle x^{r}\left[\sum_{n=0}^{\infty}\left(r^{2}+(2n-1)r+n^{2}-n+1\right)c_{n}x^{n+1}-\sum_{n=0}^{\infty}\left(r^{2}+2nr+n^{2}-1\right)c_{n}x^{n}\right]=0\)
Man, this is tedious business. It has been awhile since I delved into this stuff. But, continue by adjusting so the indices are the same.
The C's can be found by using initial conditions. r can be found when the indices are adjusted. That is, find the n=0 case.
Letting k=n+1 will aid in setting the indices equal.
In this problem, I am pretty sure we get
\(\displaystyle r^{2}-1=0\). Which means \(\displaystyle r=\pm 1\)
Then, the iterations come in.
