I've been trying to solve this one for a while and haven't had any success. Since I'm not exactly sure how formatting works, I figured I'd just scan what I have done so far and post it.
I'm not sure what to do with the terms that come before the summation, like $c_{-2}$ and $c_{-1}$. Can I ignore them after using them to solve for the other terms, given the assumption that the power series we use for $y$ starts at $n=0$?
$$y''+x^2y'+xy=0$$ $$y=\sum_{n=0}^\infty c_nx^n$$ $$\sum_{n=2}^\infty c_nn(n-1)x^{n-2}+x^2\sum_{n=1}^\infty c_nnx^{n-1}+x\sum_{n=0}^\infty c_nx^n=0$$ $$\sum_{n=2}^\infty c_nn(n-1)x^{n-2}+\sum_{n=1}^\infty c_nnx^{n+1}+\sum_{n=0}^\infty c_nx^{n+1}=0$$ $$\sum_{n=0}^\infty c_{n+2}(n+2)(n+1)x^{n}+\sum_{n=2}^\infty c_{n-1}(n-1)x^{n}+\sum_{n=1}^\infty c_{n-1}x^{n}=0$$ $$c_0x+\sum_{n=0}^\infty c_{n+2}(n+2)(n+1)x^{n}+\sum_{n=2}^\infty c_{n-1}(n)x^{n}=0$$ $$c_0x+2c_2+6c_3x+\sum_{n=2}^\infty( c_{n+2}(n+2)(n+1)+c_{n-1}n)x^{n}=0$$ Hence for $n \ge 2$: $$(c_0+6c_3)x+2c_2=0$$ $$ c_{n+2}=-\dfrac {nc_{n-1}}{(n+2)(n+1)}$$ Or more simply for $n \ge 1$:: $$c_2=0$$ $$ c_{n+2}=-\dfrac {nc_{n-1}}{(n+2)(n+1)}$$ Is this what you have ?