How to solve this partial differential equation ? (PDE) (UPDATED) How can I solve this PDE?
$$ \frac{{∂}^2z}{∂x^2} - \frac{∂^2z}{∂x∂y} - 2\frac{∂^2z}{∂y^2} +6\frac{∂z}{∂x}- 9\frac{∂z}{∂y} +5z = e^{2x +y} + e^{x+y} $$

The given equation is $(D^2-DD'-2D'^2+6D-9D'+5)z=e^{2x+y}+e^{x+y}$
The auxillary equation is obtained by setting $D=m, D'=1$ since there are no common factors. This gives $m^2-m-2+6m-9+5=0\implies m^2+5m-6=0$. Call the roots $c_1$ and $c_2$. So the homogeneous (complementary) solution is $z_c(x,y)=\phi_1(y+c_1x)+\phi_2(y+c_2x)$.
The particular integral is $$\dfrac{1}{(D^2-DD'-2D'^2+6D-9D'+5)}(e^{2x+y})=\dfrac{1}{4-2-2+12-9+5}e^{2x+y}=\dfrac{1}{8}e^{2x+y}$$
For the 2nd term the denominator becomes $0$. In that case differentiate $F(D,D')$ wrt $D$ and multiply by $x$. This means, we have
$$\dfrac{1}{(2D-D'+6)}(x\cdot e^{x+y})=\dfrac{x}{7}e^{x+y}$$
So the final solution is $z(x,y)=\phi_1(y+c_1x)+\phi_2(y+c_2x)+\dfrac{1}{8}e^{2x+y}+\dfrac{1}{7}x\cdot e^{x+y}$.
