Start by using \(\displaystyle \displaystyle y=\sum_{n=0}^{\infty}c_{n}x^{n+r}\)
. Now, find y' and y'' by taking the first and second derivatives of y above.
Sub these into your DE to get it in terms of series.
Solving using series can be rather tedious but effective method for solving.
The idea is to get your powers of x the same. That is, the series should have the same starting point.
Doing the above gives us:
\(\displaystyle \displaystyle \sum_{n=0}^{\infty}c_{n}(n+r)(n+r-1)x^{n+r}+6\)\(\displaystyle \displaystyle\sum_{n=0}^{\infty}c_{n}(n+r)x^{n+r}+\sum_{n=0}^{\infty}c_{n}x^{n+r+2}+6\sum_{n=0}^{\infty}c_{n}x^{n+r}=0\)
\(\displaystyle \displaystyle x^{r}\left[\sum_{n=0}^{\infty}\left[(n+r)(n+r-1)+6(n+r)+6\right]c_{n}x_{n}+\sum_{n=0}c_{n}x^{n+2}\right]=0\)
Now, let \(\displaystyle k=n, \;\ k=n+2\)
\(\displaystyle \displaystyle x^{r}\left[\sum_{k=0}^{\infty}\left[(k+r)(k+r-1)+6(k+r)+6\right]c_{k}x^{k}+\sum_{k=2}^{\infty}c_{k-2}x^{k}\right]=0\)
There's a start. Now, on the first sum, write out k=0 and k=1 so the indices are the same on the sums.
