How to solve this partial differential equation ? (PDE) (UPDATED) How can I solve this PDE?
$$ \frac{{∂}^2z}{∂x^2} - \frac{∂^2z}{∂x∂y} - 2\frac{∂^2z}{∂y^2} +6\frac{∂z}{∂x}- 9\frac{∂z}{∂y} +5z = e^{2x +y} + e^{x+y} $$
 A: Hint:
Note that $x^2-xy-2y^2+6x-9y$ cannot factorize in $\mathbb{R}$ according in http://www.wolframalpha.com/input/?i=factorize+x%5E2-xy-2y%5E2%2B6x-9y
Let $z=e^{ax+by}u$ ,
Then $\dfrac{\partial z}{\partial x}=e^{ax+by}\dfrac{\partial u}{\partial x}+ae^{ax+by}u$
$\dfrac{\partial^2z}{\partial x^2}=e^{ax+by}\dfrac{\partial^2u}{\partial x^2}+ae^{ax+by}\dfrac{\partial u}{\partial x}+ae^{ax+by}\dfrac{\partial u}{\partial x}+a^2e^{ax+by}u=e^{ax+by}\dfrac{\partial^2u}{\partial x^2}+2ae^{ax+by}\dfrac{\partial u}{\partial x}+a^2e^{ax+by}u$
$\dfrac{\partial^2z}{\partial x\partial y}=e^{ax+by}\dfrac{\partial^2u}{\partial x\partial y}+be^{ax+by}\dfrac{\partial u}{\partial x}+ae^{ax+by}\dfrac{\partial u}{\partial y}+abe^{ax+by}u$
$\dfrac{\partial z}{\partial y}=e^{ax+by}\dfrac{\partial u}{\partial y}+be^{ax+by}u$
$\dfrac{\partial^2z}{\partial y^2}=e^{ax+by}\dfrac{\partial^2u}{\partial y^2}+be^{ax+by}\dfrac{\partial u}{\partial y}+be^{ax+by}\dfrac{\partial u}{\partial y}+b^2e^{ax+by}u=e^{ax+by}\dfrac{\partial^2u}{\partial y^2}+2be^{ax+by}\dfrac{\partial u}{\partial y}+b^2e^{ax+by}u$
$\therefore e^{ax+by}\dfrac{\partial^2u}{\partial x^2}+2ae^{ax+by}\dfrac{\partial u}{\partial x}+a^2e^{ax+by}u-e^{ax+by}\dfrac{\partial^2u}{\partial x\partial y}-be^{ax+by}\dfrac{\partial u}{\partial x}-ae^{ax+by}\dfrac{\partial u}{\partial y}-abe^{ax+by}u-2e^{ax+by}\dfrac{\partial^2u}{\partial y^2}-4be^{ax+by}\dfrac{\partial u}{\partial y}-2b^2e^{ax+by}u+6e^{ax+by}\dfrac{\partial u}{\partial x}+6ae^{ax+by}u-9e^{ax+by}\dfrac{\partial u}{\partial y}-9be^{ax+by}u=e^{2x+y}+e^{x+y}$
$\dfrac{\partial^2u}{\partial x^2}+2a\dfrac{\partial u}{\partial x}+a^2u-\dfrac{\partial^2u}{\partial x\partial y}-b\dfrac{\partial u}{\partial x}-a\dfrac{\partial u}{\partial y}-abu-2\dfrac{\partial^2u}{\partial y^2}-4b\dfrac{\partial u}{\partial y}-2b^2u+6\dfrac{\partial u}{\partial x}+6au-9\dfrac{\partial u}{\partial y}-9bu=e^{(2-a)x+(1-b)y}+e^{(1-a)x+(1-b)y}$
$\dfrac{\partial^2u}{\partial x^2}-\dfrac{\partial^2u}{\partial x\partial y}-2\dfrac{\partial^2u}{\partial y^2}+(2a-b+6)\dfrac{\partial u}{\partial x}-(a+4b+9)\dfrac{\partial u}{\partial y}+(a^2-ab-2b^2+6a-9b)u=e^{(2-a)x+(1-b)y}+e^{(1-a)x+(1-b)y}$
Choose $2a-b+6=0$ and $a+4b+9=0$ , i.e. $a=-\dfrac{11}{3}$ and $b=-\dfrac{4}{3}$ , the PDE becomes $\dfrac{\partial^2u}{\partial x^2}-\dfrac{\partial^2u}{\partial x\partial y}-2\dfrac{\partial^2u}{\partial y^2}-5u=e^\frac{17x+7y}{3}+e^\frac{17x+7y}{3}$
A: 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 $z(x,y)=\phi_1(y+c_1x)+\phi_2(y+c_2x)+\dfrac{1}{8}e^{2x+y}+\dfrac{1}{7}x\cdot e^{x+y}$.
