Solving simultaneous partial differential equations (first-order) 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!?
$$\begin{cases} x \frac{∂f}{∂y} - y \frac{∂f}{∂x} + g = 0 \; ,\\ x \frac{∂g}{∂y} - y \frac{∂g}{∂x} - f = 0 \; .\end{cases}$$
 A: $$\begin{cases} x \frac{∂f}{∂y} - y \frac{∂f}{∂x} + g = 0 \; ,\\ x \frac{∂g}{∂y} - y \frac{∂g}{∂x} - f = 0 \; .\end{cases}$$
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 : 
$$\begin{cases}f(x,y)=x\:F(x^2+y^2)+y\:G(x^2+y^2)\\g(x,y)=y\:F(x^2+y^2)-x\:G(x^2+y^2)\end{cases}$$
A: Solution that is shorter than the one below, applying the method of characteristics directly to the problem as stated.
Introduce $x(t)$ and $y(t)$ such that $dx/dt=-y$ and $dy/dt=x$. Then the equations become
$$\begin{cases} \frac{df}{dt} + g = 0 \; ,\\ \frac{dg}{dt} - f = 0 \; .\end{cases}$$
The four equations can be solved to give
$$\begin{cases} x=a\cos t + b \sin t  \; ,\\ y=a\sin t - b \cos t \; ,\\  f=A\cos t + B \sin t  \; ,\\ g=A\sin t - B \cos t \; .\end{cases}$$
Set the initial conditions to $x(0)=x_0$, $y(0)=0$, $f(0)=u(x_0)$ and $g(0)=v(x_0)$. Then we have
$$\begin{cases} x=x_0\cos t  \; ,\\ y=x_0\sin t \; ,\\  f=u(x_0)\cos t + v(x_0) \sin t  \; ,\\ g=u(x_0)\sin t - v(x_0) \cos t \; .\end{cases}$$
But $x_0=\sqrt{x^2+y^2}$, thus we can rewrite the solutions as
$$\begin{cases} f=u(\sqrt{x^2+y^2})\frac{x}{\sqrt{x^2+y^2}} + v(\sqrt{x^2+y^2}) \frac{y}{\sqrt{x^2+y^2}}  \; ,\\ g=u(\sqrt{x^2+y^2})\frac{y}{\sqrt{x^2+y^2}} - v(\sqrt{x^2+y^2}) \frac{x}{\sqrt{x^2+y^2}} \; .\end{cases}$$

Solution using the hints in the comments. First let's introduce a new notation for the differential operator
$$H:=x \frac{∂}{∂y} - y \frac{∂}{∂x} \; .$$
The equations then become 
$$\begin{cases} Hf + g = 0 \; ,\\ Hg - f = 0 \; .\end{cases}$$
Substituting the last equation in the first, we get an equation for $g$ only
$$H^2g+g=0 \; ,$$
which we can rewrite as 
$$(H+i)(H-i)g=0$$
So the general solution for $g$ will be a linear combination of solutions for 
$$(H+i)g=0 \;\; \text{ and } \;\; (H-i)g=0$$
Let's focus our attention on the last one, the other one can be solved similarly. Let's also introduce a new variable $z=\ln(g)$, then the last equation is
$$x \frac{∂z}{∂y} - y \frac{∂z}{∂x} = i$$
We can now easily apply the method of characteristics, put
$$\begin{cases}\frac{dx}{dt}=-y \\ \frac{dy}{dt}=x \\ \frac{dz}{dt}=i\end{cases}$$
The general solutions being
$$\begin{cases}x(t)=A\cos t + B \sin t \\ y(t) = A\sin t - B \cos t \\ z(x(t),t)=it+C\end{cases}$$
We now put as initial conditions $x(0)=x_0$, $y(0)=0$ and $z(x_0,0)=u(x_0)$. This means 
$$\begin{cases}x(t)=x_0\cos t \\ y(t) = x_0\sin t \\ z(x(t),t)=it+u(x_0)\end{cases}$$
Restating that last part in terms of $g$ and introducing $v(x)=\exp{u(x)}$:
$$\begin{cases}x(t)=x_0\cos t \\ y(t) = x_0\sin t \\ g(x(t),t)=v(x_0)e^{it}\end{cases}$$
We can find out of the first equation that $x_0=\sqrt{x^2+y^2}$ and thus write
$$g(x,y)=v\left(\sqrt{x^2+y^2}\right)\frac{x+iy}{\sqrt{x^2+y^2}}$$
Solving the other equation $(H+i)g=0$, it is clear we'll get a complex conjugate of the one we just solved, which means we can always express the complete solution $g$ as a real function
$$g(x,y)=v\left(\sqrt{x^2+y^2}\right)\frac{b_1 x+b_2 y}{\sqrt{x^2+y^2}}$$
Filling this in the remaining equation
$$x \frac{∂g}{∂y} - y \frac{∂g}{∂x} - f = 0$$
We can find the formula for $f$
$$f(x,y)=v\left(\sqrt{x^2+y^2}\right)\frac{b_2 x-b_1 y}{\sqrt{x^2+y^2}}$$
Also, we can group the factors with the square root in just one function $w\left(\sqrt{x^2+y^2}\right)=v\left(\sqrt{x^2+y^2}\right)/\sqrt{x^2+y^2}$, obviously.
