Analytic solution to Poisson equation I need to find the analytic solution to this equation, in order to compare it with solution I get from using a numerical solution. However, I have not been able to find the solution. 

I think I can't use separation of variables since the R.H.S cannot be separated. I tried solving it using Green's theorem but to no luck. Any ideas how to approach this?

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$

Boundary conditions

$u(x, 0) = u(x, 1) = 0$, $x ∈ [0, 1]$

$u(0, y) = u(1, y) = 0$, $y ∈ [0, 1]$
I found the solution:

Since 

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$

The function u(x,t) must have $e^{x−y}$

From the boundary conditions:


$u(x, 0) = 0$ then u(x,t) must be 0 for x=0, so we assume u(x,t) must
have $e^{x−y}*x$
$u(x, 1) = 0$ then u(x,t) must be 0 for x=1, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)$
$u(0, y) = 0$ then u(x,t) must be 0 for y=0, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)*y$
$u(1, y) = 0$ then u(x,t) must be 0 for y=1, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)*y*(1-y)$


The solution should look like this: $u(x,t)=e^{x−y}*x*(1-x)*y*(1-y)$

Doing the calculations leads to:

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$

Which due to the uniqueness of the solution, is the solution.

Calculations can be seen below:
with free CAS Maxima:

(%i1)   load(diffops);

(%o1)   "C:\Users\aleksas\maxima\diffops.mac"

(%i2)   eq:'diff(u,x,2)+'diff(u,y,2)=2*x*(y-1)*(y-2*x+x*y+2)*exp(x-y);

(eq)    'diff(u,y,2)+'diff(u,x,2)=2*x*(y-1)*(x*y+y-2*x+2)*%e^(x-y)

(%i3)   partsolpde(eq,u,[x,y]);

(%o3)   x^2*y^2*%e^(x-y)-x*y^2*%e^(x-y)-x^2*y*%e^(x-y)+x*y*%e^(x-y)

(%i4)   factor(%);

(%o4)   (x-1)*x*(y-1)*y*%e^(x-y)


This is solution