Given is $\sum\limits_{k=0}^\infty a_kx^k$ and $f''(x)-xf(x)=x$, $f'(0)=0$, $f(0)=0$. \newline I need to find a series representation of the upper sum, through determining an expression for $a_k$ using the given relations. I know how to solve such an expression when $f''(x)-xf(x)=x$ is instead $f''(x)-xf(x)=0$, but if I remove the x here I get $\frac{f''(x)}{x}-f(x)=1$ and I am not sure how to solve that, as I do not get a logical answer when I tried. So how can I solve this? I got: $\sum\limits_{k=1}^\infty (k+2)(k+3)a_{k+3}x^{k}-\sum\limits_{k=3}^\infty a_{k}x^{k}=1$
But how do I need to work with that last 1 in order to be able to work with the identity theorem and thus obtain a reursive formula?`
This is a linear differential equation, and one solution of it is $f(x) = -1$. The general solution is then $f(x) = g(x) - 1$ where $g$ is a solution of the homogeneous equation $g''(x) - x g(x) = 0$. The given initial conditions on $f$ gives you $g(0) = 1$, $g'(0)=0$. Now find a power series for $g$.