how to bring the PDE $u_{tt}-u_{xx} = x^2 -t^2$ to the canonical form How to bring to the canonical form and solve the below PDE?
$$u_{tt}-u_{xx} = x^2 -t^2$$
I recognize that it is a hyperbolic PDE, as the $b^2-4ac=(-4(1)(-1))=4 > 0$.
I don't know how to proceed further to get the canonical form.
I know how to deal with something like $u_{tt}-u_{xx} = 0$.
With $\ RHS =0 \ $ I would use the equation for characteristic 
$R (\frac{\partial^2 dy}{\partial dx})-2S (\frac{\partial^2 dy}{\partial dx})+T=0$ , define the $\xi$ and $\eta$ in terms of $x$ and $y$, calculate the first and second partial derivatives and substitute them into the initial equation.
Here the function on the right hand side $x^2 -t^2$ complicates matter.
How the RHS=X^2-t^2 changes the standard wave equation $u_{tt}−u_{xx}=0$ in terms of interpretation?

Summarizing the solution should be 
the canonical form $\mathbf{-4u_{\xi \eta}=\xi \eta}$ 
the general solution $\mathbf{u(x,y)=-\frac{1}{16}(x^4-2x^2t^2+t^4)}$ 
$\mathbf{x^2-t^2}$ represent an external force
