Solving second order non-homogenous PDE $$ d_1u_{xx} +d_2 u_{yy} = -2 $$ 
I need to solve this given PDE. I tried to solve it using change of variables. The variables are 
$$ \xi = y-ax \ , \ \ \ \eta = y+ax$$  where $$a = \sqrt{-d_1d_2} $$
After the change of variable $(x,y)$ into $(\xi,\eta)$ , my equation looks like this,
$$u_{\xi\xi}(1-d_1d_2) + 2 u_{\xi\eta} (1+d_1d_2) + u_{\eta\eta}(1-d_1d_2) = -2 $$
which seems more complicated than the original equation. Did i do something wrong? When I tried with example equations I was able to solve them without any problem. Can someone please give me a clue on how to solve this one. Thanks in advance.

The problem is easier than you were trying to make it.  Let 
$$
\xi = \frac{x}{\sqrt{d_1}}, \eta = \frac{y}{\sqrt{d_2}}
$$
then the equation is 
$$
\nabla^2 u = -2
$$
and the solution to that is any solution to 
$$
\nabla^2 u = -0
$$
(all of which are of the form $u = pxy + ax+by+c$ if you want the equation to hold everywhere on the $(x,y)$ plane)
plus a particular solution to 
$$
\nabla^2 u = -2
$$
which has the family of solutions 
$$
u = m \xi^2 -(2+m) \eta^2
$$ 
for any real $m$.
So your answer should be
$$
u = \frac{m}{d_1}x^2 - \frac{2+m}{d_2} y^2 + pxy + ax + by + c  
$$
If, on the other hand, the problem is a boundary value problem one some closed subsset of the plane, there are further solutions to the homogneuous part $\nabla^2 u = 0$ which can be thought of as potentials of an arbitrary charge distribution.
