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

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

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

Then I found the first and second derivatives.

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

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

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

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

Ok, I understand up to this part. Now I'm just confused after this, but this is what I got. I took out $\displaystyle x^r$

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

Then I tried to get $\displaystyle a_0$ and changed some of the n values to $\displaystyle j $and $\displaystyle j+1$ similar to what my professor did.

$\displaystyle x^r [(2r(r-1)+3r-1)a_0 $ + $\displaystyle \sum_{n=1}^{\infty} $$\displaystyle 2(j+r)(j+r-1)a_j + 3(j+r)a_j - a_j - (j-1+r)a_{j-1}]x^j$

Then I tried to find the r by:

$\displaystyle 2r^2 +r - 1 = 0$

$\displaystyle (2r-1)(r+1)$

I got r = 1/2 and r = -1.

Is this correct so far? When my professor went over a homework question similar to this problem, he said that the r's would be -1 and -1/2. I just don't see where the -1/2 came from and even so. When I plug -1 , I get zero in the third term which ruins it unless that's ok?

I know what to do pass this part, but it's very confusing. I spent more than 5 hours trying to do this and I know I'm doing something wrong or thinking of something else. Please help me lead to the right direction.