Singular points and finding linearly independent solutions

The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$

1. How would I go proving a singular point that is regular?

I have the textbook in front of me and it doesn't help me at all. I was able to find x=0, but the question asks me to show it. So I did by making y'' equaling to 1 and divided the variable to the rest. I'm just confused if that's good enough. In my end, I don't think it is because they say show that it's a regular singluar point. I'm just a bit confused if I should show more or if what I just mentioned is ok.

2. Find two linearly independent solutions of the form $\displaystyle y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$. Using the recurrence relation.

Ok so I changed the given y to $\displaystyle y = \sum_{n=0}^{\infty} a_nx^{n+r}$

Then I found the first and second derivatives.

$\displaystyle f'(x) = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1}$

$\displaystyle f''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2}$

When I plug it in back into the original equation, I get:

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

Am I doing this correct so far? I would like to make sure before I continue and would it be easier to take out $\displaystyle x^r$,$\displaystyle x^{r-1}$, or does it not matter?

Thanks. I'll come back after seeing if I'm doing the first part correct and post my solution to see if y'all agree.