Second order inhomogeneous PDE I'm trying to get an exact solution to this second order inhomogeneous PDE:
$$
\frac{\partial^2}{\partial{x}^2} y(x, z) - \frac{\partial^2}{\partial z^2} y(x, z)=k^2y(x, z)-\frac{1}{3}e^{4(x-2z)}y(x, z)
$$
where $k^2$ is a constant. No boundary conditions.
Any ideas? I tried with $t=(x-cz)$ and with variable separation but it isn't the right way.
Yesterday I shared the same question on Mathematics, maybe I'm luckier here
 A: I am writing my comment as answer. If we change the variables $x,z \rightarrow u,v$ such that $u=x-2y$ (Just to make the exponential term a single variable function), and $v=ax+by$, then
Then, to eliminate $\frac{\partial^2 y}{\partial u\partial v}$ term we need to choose $\frac{a}{b}=-2$. Choose, $a=-2, b=1$.
Then, the equation becomes $\frac{\partial ^2 y}{\partial v^2}+\frac{k^2}{3}y=\frac{\partial ^2 y}{\partial u^2}+\frac{1}{9}{e^{4u}}y$.
Separating $y=B(v)A(u)$ we get, $B''(v)+(\frac{k^2}{3}-b)B(v)=0$ for some constant which communicates between two sides. And this is easy to solve.
The second equation is $A''(u)+(\frac{1}{9}e^{4u}-b)A(u)=0$.
Now ,again change the variable as $u \rightarrow t=e^{2u}$, then the equation becomes $u^2\bar{A}''+u\bar{A}'+((\frac{t}{6})^2-(\frac{\sqrt{b}}{2})^2)\bar{A}=0$ where, $\bar{A}(t)=A(u)$.
And this is the Bessel equation, hence, $A(u)=c_1J_{\frac{\sqrt{b}}{2}}(\frac{e^{2u}}{6})+c_2J_{-\frac{\sqrt{b}}{2}}(\frac{e^{2u}}{6})$.
($\frac{\sqrt{b}}{2}$ isn't a natural number, in that case Bessel function of second kind should be taken).
