I have to solve
$$y''+y'-2y=0$$ with $y(0)=0$ and $y'(0)=1$.
Using $$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$ I got $a_{0}=0,a_{1}=1$ and $$a_{n}=\frac{(n+1)(a_{n+1}+(n+2)a_{n+2})}{2}$$ And I really don't know what can I do now. I mean: is there some recursive thing? I calculated the value of $a_{2},a_{3},a_{4},a_{5}$ and $a_{6}$ and I couldn't see anything.
Beside the solution suggested by Brevan Ellefsen, you could rewrite $$a_{n}=\frac{(n+1)(a_{n+1}+(n+2)a_{n+2})}{2}$$ as $$2a_n=(n+1)a_{n+1}+(n+1)(n+2)a_{n+2}$$ Muliply each side by $n!$ and simplify to get $$2\,n!\, a_n=(n+1)!\,a_{n+1}+(n+2)!\, a_{n+2}$$Now, define $b_n=n! \, a_n$ to get $$2 b_n=b_{n+1}+b_{n+2}$$ for which the characteristic equation is $$2=r+r^2\implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.