Hyperbolic non-homogeneous 2nd order linear PDE Solve the following PDE:
$$u_{xy} - 2u_{x} - 3u_y + 6u=2e^{xy}$$ 
I've recently learnt how to bring pde to their canonical form. This one, as far as I can tell, it ia already in canonical form. It is a hyperbolic equation. But how am I supposed to solve it? 
I googled it and found that I can look for solutions of the form $u = X(x)Y(y)$ and then separate variables. But it doesn't seem to work in this particular example. Both because I have some free terms (non-homogeneity) and bcs I well....it's complicated. Any guidances? 
 A: You may solve this equation using the method of order reduction.
Let us begin from a re-expression of the original equation. Note that
\begin{align}
\text{LHS}&=\left(\partial_y-2\right)u_x-3\left(\partial_y-2\right)u\\
&=\left(\partial_y-2\right)\partial_xu-3\left(\partial_y-2\right)u\\
&=\partial_x\left(\partial_y-2\right)u-3\left(\partial_y-2\right)u\\
&=\left(\partial_x-3\right)\left(\partial_y-2\right)u.
\end{align}
Therefore, the original equation reads
$$
\left(\partial_x-3\right)\left(\partial_y-2\right)u=e^{xy},
$$
which can be split into
\begin{align}
\left(\partial_y-2\right)u&=v,\\
\left(\partial_x-3\right)v&=e^{xy}.
\end{align}
Now, each of the equations above can be regarded as an ordinary differential equation, as only one partial derivative appears.
For the second equation, we have
$$
\left(\partial_x-3\right)v=e^{xy}\iff\partial_x\left(e^{-3x}v\right)=e^{-3x}e^{xy}=e^{x\left(y-3\right)}.
$$
Therefore,
$$
e^{-3x}v(x,y)-v(0,y)=\int_0^x\partial_x\left(e^{-3s}v(s,y)\right){\rm d}s=\int_0^xe^{s\left(y-3\right)}{\rm d}s=\frac{e^{x\left(y-3\right)}-1}{y-3}.
$$
Consequently,
$$
v(x,y)=e^{3x}v(0,y)+\frac{e^{xy}-e^{3x}}{y-3}.
$$
With this result, the first equation yields
$$
\left(\partial_y-2\right)u=v\iff\partial_y\left(e^{-2y}u\right)=e^{-2y}v=e^{3x-2y}v(0,y)+\frac{e^{\left(x-2\right)y}-e^{3x-2y}}{y-3}.
$$
A straightforward integration gives
$$
e^{-2y}u(x,y)=u(x,0)+e^{3x}\int_0^ye^{-2s}v(0,s){\rm d}s+\int_0^y\frac{e^{\left(x-2\right)s}-e^{3x-2s}}{s-3}{\rm d}s,
$$
or equivalently,
$$
u(x,y)=e^{2y}u(x,0)+e^{3x+2y}\int_0^ye^{-2s}v(0,s){\rm d}s+e^{2y}\int_0^y\frac{e^{\left(x-2\right)s}-e^{3x-2s}}{s-3}{\rm d}s.
$$
The last integration from above could unfortunately not be simplified in an easy fashion, but this term is independent from the unknowns. Plus, the choice of $u(x,0)$ and $v(0,y)$ should be such that your boundary conditions are satisfied.
