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.
\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}