Help needed in solving a differential equation Please help me in solving:
$$a^2z+\frac{\partial^2z}{\partial x^2}-\frac{\partial^2 z}{\partial y^2}=0$$ 
($a$ is a constant)
I plugged this in Wolfram Alpha and it outputs that this is a second order linear partial differential equation but unfortunately, I have never solved these type of equations before. I came across this equation while solving some other problem. I tried to look up the methods to solve such equations but they seem too advanced to me. :( Can somebody please please solve this? I am not even sure if it is possible to solve this equation.
Any help is appreciated. Thanks!
 A: Suggestion (for a domain bounded in $x$):
Try separation of variables, $z(x,y) = X(x)Y(y)$, leading to two ODE eigenvalue problems
$$X''(x) = cX(x) \\\ Y''(y) =(c+\alpha^2)Y(y)$$
For a semi-infinite domain ($0 \leq x<\infty,0 \leq y<\infty$) we have a wave equation
$$z_{yy}-z_{xx}= a^2z.$$
If initial/boundary conditions are correctly specified -- making the problem well-posed --then a method for solution is to take the Laplace transform with respect to $x$ and solve the resulting second-order ODE for the tranform.
A: Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,
Let $\begin{cases}p=\dfrac{x+y}{2}\\q=\dfrac{x-y}{2}\end{cases}$ ,
Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)$
$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)$
$\dfrac{\partial^2z}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial x}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$
$\dfrac{\partial^2z}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial y}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$
$\therefore a^2z+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=0$
$a^2z+\dfrac{\partial^2z}{\partial p\partial q}=0$
$\dfrac{\partial^2z}{\partial p\partial q}=-a^2z$
Similar to Second order hyperbolic PDE,
The general solution is $z(p,q)=\int_0^pf(s)J_0\left(\dfrac{2\sqrt{q(p-s)}}{a}\right)~ds+\int_0^qg(s)J_0\left(\dfrac{2\sqrt{p(q-s)}}{a}\right)~ds$
$z(x,y)=\int_0^\frac{x+y}{2}f(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x-y}{2}\left(\dfrac{x+y}{2}-s\right)}\right)~ds+\int_0^\frac{x-y}{2}g(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x+y}{2}\left(\dfrac{x-y}{2}-s\right)}\right)~ds$
$z(x,y)=\int_0^{x+y}F(s)J_0\left(\dfrac{\sqrt{(x-y)(x+y-s)}}{a}\right)~ds+\int_0^{x-y}G(s)J_0\left(\dfrac{\sqrt{(x+y)(x-y-s)}}{a}\right)~ds$
