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.
 A: 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.
A: Just for completeness,
the discriminant of your PDE is given by: 
$$B^2 - 4AC = 0 - 4 d_1 d_2, \quad A = d_1, B = 0, C=d_2, $$ which can be positive, negative or zero depending on the values of $d_i$. Since, a priori, we don't known anything of the coefficients $d_i$, we have that the characteristics of your PDE can be found from the following equation:
$$d_1 \xi_{x}^2 + d_2 \xi_{y}^2 = 0,  \tag{1}$$ (and the same for the other characteristic, $\eta$) which can be solved for:
a) $d_{1,2} \neq 0$: which yields: $\xi_x/\xi_y = \pm \sqrt{d_2/d_1} = - dy/dx$, giving the characteristics curves:
$$y + \alpha x = \xi, \quad y - \alpha x = \eta, \quad \alpha = \sqrt{d_2/d_1}. $$ After using the chain rule, we obtain the original equation written in terms of the new variables:
$$u_{\xi\xi} + u_{\eta \eta} = \frac{1}{d_2}, \tag{2}$$
which is an elliptic PDE (Poisson's eq.) you can solve following @Mark Fischler's answer.
b) $d_1 = 0, d_2 \neq 0$: we would have $d_2 u_{yy} = -2$, whose general solution is:
$$u(x,y) = - \frac{y^2}{d_2}  + y f(x) + g(x), \tag{3}$$ being $f$ and $g$ arbitrary functions of $x$.
c) $d_2 = 0, d_1 \neq 0$: reasoning as we have done above, we would get:
$$u(x,y) = - \frac{x^2}{d_1} + x f(y) + g(y). \tag{4}$$
Hope this helps!
Cheers.
