finding explicit solution to PDE with given initial conditions A solution to a PDE of my interest is
$u(x,y)=2\left(F''(x)-G''(y)\right)(x-y)^{2}-12\left(F'(x)+G'(y)\right)(x-y)+24\left(F(x)-G(y)\right)$
where $F(x)$ and $G(y)$ are arbitrary functions to be determined from initial conditions. In this case, I know
$u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}}$,
$u(x_{0},y)=\frac{u_{0}(x_{0}-y)^{2}}{(x_{0}-y_{0})^{2}}$.
I tried all sorts of things to find $F(x)$ and $G(y)$. I tried all sorts of things to no avail; any suggestions what steps/approaches I should take?
 A: I made some progress on this so I thought I would share it. First let $u(x_{0},y_{0})=u_{0}$ then one can write the following
$
u(x_{0},y_{0})=2\left(F''(x_{0}-G''(y_{0})\right)(x_{0}-y_{0})^{2}-12\left(F'(x_{0})+G'(y_{0})\right)(x_{0}-y_{0})+24\left(F(x_{0}-G(y_{0}\right)=u_{0}
$
Now choose five values arbitrarily and the 6th value will be determined.
I made the following choices:
$F''(x_{0})=1\quad F'(x_{0})=1\quad F(x_{0})=1 \quad
G''(y_{0})=?\quad  G'(y_{0})=-1 \quad G(y_{0})=1$
Then find $G''(y_{0})$ as follows:
$
u_{0}=2(1-G''(y_{0}))(x_{0}-y_{0})^{2}
$
and this gives $G''(y_{0})=1-k$ where
$
k=\frac{1}{2}\frac{u_{0}}{(x_{0}-y_{0})^{2}}
$
Now apply the initial conditions. Let's start with: 
$
u(x,y_{0})=\frac{u_{0}(x-y_{0})^{2}}{(x_{0}-y_{0})^{2}}
$
Plugging this into the solution of the pde, using what we have abve and simplifying we obtain
$
(x-y_{0})^{2}F''(x)-6(x-y_{0})F'(x)+12F(x)=H(x)
$
where
$H(x)=(x-y_{0})^{2}+6(x-y_{0})+12 $
Now what we have a second order non-homogeneous ODE and to solve for F first compute the fundamental solutions and proceed via the route of variation of parameters. One will use the other initial condition to find G. I will make an update soon on the full solution.
