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?

$$u_{tt}-u_{xx}=x^2-t^2$$
An obvious particular solution is $u=\frac{1}{12}(-x^4-t^4)$
The change of function : $u(x,t)=v(x,t)+\frac{1}{12}(-x^4-t^4)$ leads to :
$$v_{tt}-v_{xx}=0$$
$$v(x,t)=F(x+t)+G(x-t)$$
$F$ and $G$ are any derivable functions.
$$u(x,t)=F(x+t)+G(x-t)-\frac{1}{12}(x^4+t^4)$$
