Solutions of $x\frac{\partial \psi}{\partial x} + y \frac{\partial \psi}{\partial y} + \psi = f(x)e^{-2\pi i y}$?

Question

I stumbled across the following PDE for a function $\psi(x, y)$: $$ x\frac{\partial \psi}{\partial x} + y \frac{\partial \psi}{\partial y} + \psi = f(x)e^{-2\pi i y} $$ where $f(z)$ is some arbitrary function. I was wondering if anyone knows anything about equations of this kind or how to solve them. Separation of variables didn't seem to work, so I'm fresh out of ideas.

Answer 1

Maple says $$ \psi(x,y) = \frac{1}{x}\left(\int_a^xf(t)\exp\left(\frac{-2\pi i y t}{x}\right)\;dt+F\left(\frac{y}{x}\right)\right) $$ where $F$ is an arbitrary function.

Answer 2

$x\dfrac{\partial\psi}{\partial x}+y\dfrac{\partial\psi}{\partial y}+\psi=f(x)e^{-2\pi iy}$

$x\dfrac{\partial\psi}{\partial x}+y\dfrac{\partial\psi}{\partial y}=f(x)e^{-2\pi iy}-\psi$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0x$

$\dfrac{d\psi}{dt}=f(x)e^{-2\pi iy}-\psi=f(e^t)e^{-2\pi iy_0e^t}-\psi$

$\dfrac{d\psi}{dt}+\psi=f(e^t)e^{-2\pi iy_0e^t}$

I.F. $=e^{\int dt}=e^t$

$\therefore\dfrac{d}{dt}(e^t\psi)=f(e^t)e^{t-2\pi iy_0e^t}$

$e^t\psi=\int f(e^t)e^{t-2\pi iy_0e^t}~dt$

$e^t\psi=\int_0^tf(e^s)e^{s-2\pi iy_0e^s}~ds+C$

$\psi=e^{-t}\int_0^tf(e^s)e^{s-2\pi iy_0e^s}~ds+Ce^{-t}$

$u(0)=F(y_0)$ :

$C=F(y_0)$

$\therefore\psi(x,y)=e^{-t}\int_0^tf(e^s)e^{s-2\pi iy_0e^s}~ds+F(y_0)e^{-t}$

$\psi(x,y)=\dfrac{1}{x}\int_0^{\ln x}f(e^s)e^se^{-\frac{2\pi iye^s}{x}}~ds+\dfrac{1}{x}F\left(\dfrac{y}{x}\right)$

$\psi(x,y)=\dfrac{1}{x}\int_0^{\ln x}f(e^s)e^{-\frac{2\pi iye^s}{x}}~d(e^s)+\dfrac{1}{x}F\left(\dfrac{y}{x}\right)$

$\psi(x,y)=\dfrac{1}{x}\int_1^xf(t)e^{-\frac{2\pi iyt}{x}}~dt+\dfrac{1}{x}F\left(\dfrac{y}{x}\right)$