Solving the ODE $y^{\prime\prime}(x)-y(x)=g(x)$ using the Fourier transform, without missing solutions

Question

I'm supposed to solve the ODE $y^{\prime\prime}(x)-y(x)=g(x)$ using the Fourier transform and then explain if I got the most general solution.

First of all, I don't know what "solve" means here because the furthest I can get is $$-\frac{\hat g(\omega)}{1+\omega^2}=\hat y(\omega)$$ which by the convlution theorem tells me $y=-(g\ast \frac 12 e^{-|t|})$ and I don't see what more I can do.

Second, I don't understand what solutions I am missing... Help!

Answer 1

You have a solution: $$ -\frac{1}{2}\int_{-\infty}^{\infty}g(t)e^{-|t-x|}dt $$ That solution works for a large class of functions $g$, but it is not the most general solution because you can add solutions of $y''-y=0$. So a more general solution is $$ y(x)= A e^{x}+Be^{-x}-\frac{1}{2}\int_{-\infty}^{\infty}g(t)e^{-|t-x|}dt, $$ where $A$ and $B$ are arbitrary constants.