I am looking for a function $V(x,y):\mathbb{R}\times(0,k)\to\mathbb{R}$, for some $k>0$, to solve the PDE
$$\rho V=y^2(\frac{1}{2}V_{xx}-V_y)+c$$
For some $c\neq 0$ and $\rho>0$. I am not very familiar with the PDE techniques, but my instinct was to try and solve the homogeneous case, try to find a particular solution and add them up together. I'm not sure if that works in this case, but here it is what I've got:
My Work: Consider the homogeneous PDE: $\rho V=y^2(\frac{1}{2}V_{xx}-V_y)$. This equation can be solved by separation of variables by setting $V(x,y)=X(x)Y(y)$, so that $V_{xx}=\frac{X''}{X}V$, $V_y=\frac{Y'}{Y}V$, and plugging back into the equation,
$$\rho=y^2(\frac{1}{2}\frac{X''}{X}-\frac{Y'}{Y})$$
Or
$$2\frac{\rho}{y^2}+2\frac{Y'}{Y}=\lambda=\frac{X''}{X}$$
For some constant $\lambda$. Suppose that we have boundary conditions so that we only care about the exponential solution of $X$'s second order ODE. With this, have the solution$$V(x,y)=\alpha\exp(\sqrt{\lambda}x+\frac{\lambda}{2}y+\frac{\rho}{y})+\beta\exp(-\sqrt{\lambda}x+\frac{\lambda}{2}y+\frac{\rho}{y})$$
For some $\alpha$ and $\beta$ constants and $\lambda>0$ constant as well.How should I proceed to get the solution of the general nonhomogeneous equation with $c\neq 0$ (setting aside the rest boundary conditions, which solve a free-boundary problem)?
Best Answer
Your equation is linear so your idea is completely justified. All you need is a particular solution of the non homogeneous equation. In this case the constant $V=c/\rho$ would do the job.