Q:

Transform the equation $(x+y)\frac{\partial u}{\partial x} - (x-y)\frac{\partial u}{\partial y} = 0$ with $s = \ln\sqrt{x^2+y^2}$, $t = \arctan(y/x)$

Let $u(x, y)$ be a $C^1$-smooth function.  Transform the equation
\begin{equation}
(x+y)\frac{\partial u}{\partial x} - (x-y)\frac{\partial u}{\partial y} = 0
\end{equation}
by introducing the new variables $s = \ln\sqrt{x^2+y^2}$, $t = \arctan(y/x)$. Use this to write down the general solution to the equation. 
I've tried multiple things with no luck as I think I'm way off.  I really have no idea how to rewrite the equation in terms of $t$ and $s$. Would love some help, thanks. 

A:

\begin{align}
  \frac{\partial u}{\partial x} &=
  \frac{\partial u}{\partial s} \frac{\partial s}{\partial x}+
  \frac{\partial u}{\partial t} \frac{\partial t}{\partial x} \\
  &= \frac{x}{x^2+y^2}\frac{\partial u}{\partial s}-
     \frac{y}{x^2+y^2}\frac{\partial u}{\partial t} \\
  \frac{\partial u}{\partial y} &=
  \frac{\partial u}{\partial s} \frac{\partial s}{\partial y}+
  \frac{\partial u}{\partial t} \frac{\partial t}{\partial y} \\
  &= \frac{y}{x^2+y^2}\frac{\partial u}{\partial s}+
     \frac{x}{x^2+y^2}\frac{\partial u}{\partial t} \\
  (x+y)\frac{\partial u}{\partial x}-(x-y)\frac{\partial u}{\partial y} &=
  \frac{(x+y)x-(x-y)y}{x^2+y^2} \frac{\partial u}{\partial s}+
  \frac{(x+y)(-y)-(x-y)x}{x^2+y^2} \frac{\partial u}{\partial t} \\
  &= \frac{\partial u}{\partial s}-\frac{\partial u}{\partial t}
\end{align}

