I am trying to find the general solution of $x^{2} u_{xx} -2xy u_{xy} + y^{2} u_{yy} = e^{x}$. The equation is parabolic as the only root of the characteristic equation is $\lambda = \frac{y}{x}$. Changing variables to $\xi (x,y) = xy, \eta (x, y) = y$, one gets that the canonical form of the aforementioned PDE is
$ v_{\eta \eta} - \frac{2 \xi}{\eta ^2} v_{\xi} - \frac{1}{\eta ^2} e^{\frac{\xi}{\eta}} = 0. $
I know it's possible to find an explicit expression for the general solution, but I don't know how to do it. I have tried to approach the problem by finding first the general solution of the homogeneous case (taking the $e^{\frac{\xi}{\eta}}$ out of the equation), but I still don't know what to do.
:(
We could try to separate variables with $u(x,y)=f_1(x)+f_2(y)$
This leads to $$x^{2} \left(\frac{d^{2}}{d x^{2}}\mathit{f_1}\! \left(x\right)\right)+y^{2} \left(\frac{d^{2}}{d y^{2}}\mathit{f_2}\! \left(y\right)\right)={\mathrm e}^{x}$$
This equation can be solved by the identities: $$\frac{d^{2}}{d x^{2}}f_{1}\! \left(x\right)=\frac{k}{x^{2}}+\frac{{\mathrm e}^{x}}{x^{2}}$$
$$\frac{d^{2}}{d y^{2}}f_{2}\! \left(y\right)=-\frac{k}{y^{2}}$$
Verify: $$x^{2}\cdot \left(\frac{k}{x^{2}}+\frac{{\mathrm e}^{x}}{x^{2}}\right)+y^{2}\cdot \left(-\frac{k}{y^{2}}\right)=e^{x}$$
Now we can solve both separated ODEs:
$$\mathit{f_1}(x)= \mathrm{Ei}\! \left(x\right)\cdot \left(x-1\right)-{\mathrm e}^{x}-k\cdot \ln\! \left(x\right)+c_{1}\cdot x+c_{2}$$
$$f_{2}(y)= k\cdot\ln\left(y\right)+C_{1} y+C_{2}$$
HINT: $Ei(z)$ is the exponential integral.
Finally we get
$$u(x,y)=\mathrm{Ei}\! \left(x\right)\cdot \left(x-1\right)-{\mathrm e}^{x}+c_{1}\cdot x+c_{2}+k\cdot \ln\! \left(\frac{y}{x}\right)+C_{1}\cdot y+C_{2}$$
$[k, c_1, c_2, C_1, C_2] \in \mathbb{R}$ are arbitrary constants.
PS I'm not pretty sure if this is the most general solution.