$\displaystyle y''+3y'+(1-x)y=0$

I used these:

$\displaystyle y=\sum_{n=0}^{\infty} a_{n}x^{n}$

$\displaystyle y'=\sum_{n=1}^{\infty} (n)(a_{n})x^{n-1}$

$\displaystyle y''=\sum_{n=2}^{\infty} (n)(n-1)(a_{n})x^{n-2}$

I Substitued the values in the origional equation:

$\displaystyle \sum_{n=2}^{\infty} (n)(n-1)(a_{n})x^{n-2}+3\sum_{n=1}^{\infty} (n)(a_{n})x^{n-1}+(1-x)\sum_{n=0}^{\infty} a_{n}x^{n}=0$

Next, I multiplied the $\displaystyle 3$ in, and the $\displaystyle (1-x)$:

$\displaystyle \sum_{n=2}^{\infty} (n)(n-1)(a_{n})x^{n-2}+

\sum_{n=1}^{\infty} (3n)(a_{n})x^{n-1}+

\sum_{n=0}^{\infty} a_{n}x^{n}-

\sum_{n=0}^{\infty} a_{n}x^{n+1}=0$

Now, I need all of these with the same starting value for $\displaystyle n$ and have $\displaystyle x$ to the same power.

*Here is where i get messed up*

$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1)(a_{n+2})x^{n}+

\sum_{n=0}^{\infty} (3)(n+1)(a_{n+1})x^{n}+

\sum_{n=0}^{\infty} a_{n}x^{n}-

\sum_{n=0}^{\infty} a_{n}x^{n+1}=0$

The last term in is sum has the $\displaystyle x^{n+1}$ not $\displaystyle x^{n}$, so did I do this wrong or how do I find the recursive relation?

Questions:

How do i find the recursive relation?

What is the recursive relation?

This is my FIRST time trying to use power series with differential equations so in the explanation if you could explain everything with a lot of detail(not skipping steps) that would be great.

EDIT: I used http://www.stewartcalculus.com/data/...lveDEs_Stu.pdf for reference.