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?