Frobenius equation w/ second-order root

I am given this: $y'' + \frac{2}{x}y' + y = 0$

Using Frobenius' Theorem, I was able to get the equation into this: $r(r+1)a_{0} + (r+1)(r+2)a_{1} + \sum\limits_{k = 1}^\infty[(k+r+3)(k+r+2)a_{k+2} + a_{k} = 0$

Obviously, for the $a_{0}$ term, I have to find the roots, which are -1 and 0. But what about the $a_{1}$ term? Do I need the -1 and -2 roots from there as well? Will I have 3 separate solutions by the end, or just 2?