I'm stuck on this differential equation \begin{cases} y''(x,k)+(k^2-q(x))y(x,k)=0\\ y(0,k)=0,\quad y'(0,k)=1 \end{cases} I try to integrate it directly, but it doesn't work. And there is a hint saying using variation of parameter.
In my mind, variation of parameter is feasible for $y''(x,k)+p(x)y(x,k)=0$, it seems not to be in this form or I just regard $(k^2-q(x))=p(x)$?