Solve the simultaneous equations (1 and 2) for $f(x,y)$ and $g(x,y)$. I have never seen an example like this, what is the method!?
$$\begin{cases} x \frac{∂f}{∂y} – y \frac{∂f}{∂x} + g = 0 \; ,\\ x \frac{∂g}{∂y} – y \frac{∂g}{∂x} – f = 0 \; .\end{cases}$$
Best Answer
$$\begin{cases} x \frac{∂f}{∂y} - y \frac{∂f}{∂x} + g = 0 \; ,\\ x \frac{∂g}{∂y} - y \frac{∂g}{∂x} - f = 0 \; .\end{cases}$$ The change of Cartesian system of coordinates to polar :$\quad\begin{cases}x=\rho\cos(\theta)\\y=\rho\sin(\theta)\end{cases}$
leads to a very simple system : $$\begin{cases}\frac{∂f}{∂\theta}+g=0\\ \frac{∂g}{∂\theta}-f=0\end{cases}\quad\to\quad \frac{∂^2f}{∂\theta^2}+f=0$$ $$\begin{cases}f=c_1(\rho)\cos(\theta)+c_2(\rho)\sin(\theta)\\g=-\frac{∂f}{∂\theta}=c_1(\rho)\sin(\theta)-c_2(\rho)\cos(\theta)\end{cases}$$
$\frac{c_1(\rho)}{\sqrt{x^2+y^2}}$ and $\frac{c_2(\rho)}{\sqrt{x^2+y^2}}$ can be considered as arbitrary functions of $\rho$, say $F(\rho)$ and $G(\rho)$.
Back to Cartesian coordinates : $$\begin{cases}f(x,y)=x\:F(x^2+y^2)+y\:G(x^2+y^2)\\g(x,y)=y\:F(x^2+y^2)-x\:G(x^2+y^2)\end{cases}$$