May I know how to solve the following partial differential equation?
$\displaystyle\frac{\partial z}{\partial x}=A\displaystyle\frac{\partial z}{\partial y}+B\displaystyle\frac{\partial^2 z}{\partial y^2}+Cz+D$,
where $z=z(x,y)$ is a function of $x$ and $y$; $A, B, C, D$ are constants.
If boundary conditions are needed, here they are:
$z(x,\pm\infty)=0,x\geq 0$.
Many thanks!!
The trial solution of the homogenous equation $\displaystyle\frac{\partial z}{\partial x}=A\displaystyle\frac{\partial z}{\partial y}+B\displaystyle\frac{\partial^2 z}{\partial y^2}+Cz$ has the form $z_h=Ee^{ax+by}$ and hence $$aEe^{ax+by}=AbEe^{ax+by}+Bb^2Ee^{ax+by}+CEe^{ax+by} \\ \{Ab+Bb^2+C-a\}Ee^{ax+by}=0 \\ Ab+Bb^2+C-a=0\implies b=f(a),b=g(a)\\ $$ For example if $b=2a,3a$, then $$z_h=E_1e^{ax+2ay}+E_2e^{ax+3ay}=F(x+2y)+G(x+3y)$$
The particular solution of the non-homogenous equation $\displaystyle\frac{\partial z}{\partial x}=A\displaystyle\frac{\partial z}{\partial y}+B\displaystyle\frac{\partial^2 z}{\partial y^2}+Cz+D$ has the general form of the term "$D$" i.e. $z_p=F$ is a constant and so $$0=0+0+CF+D \implies F=\frac{-D}{C}$$ The general solution is $$z=z_h+z_p$$