How to solve this PDE with method of characteristics? I have this problem $$y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1,\\u(x,0)=0$$
Using the method of characteristics I have
$$\frac{dx}{dt}=y \\ \frac{dy}{dt}=-x \\ \frac{du}{dt}=1$$ 
Then $$\frac{dx}{y}=\frac{dy}{-x} \\ x^2+y^2=\eta $$
and 
$$u=t+\xi$$
But I do not understand how solve this problem.

Write $x(t)=a\sqrt{\eta}\cos t$ and $y(t)=b\sqrt{\eta}\sin t$ where $a,b\in\{-1,+1\}$. Then $\frac{y(t)}{x(t)}=\pm\frac{\sin t}{\cos t}$ and so $t=\pm\arctan\frac{y}{x}$. If you substitute this in $u$ you get $u(x,y)=\pm\arctan\frac{y}{x}+\xi$. Now $u(x,0)=\arctan 0+\xi=0$ and so $\xi=0$. It follows that $u(x,y)=\pm\arctan\frac{y}{x}$. To determine the sign, note that
$$\frac{\partial u}{\partial x}=\pm\frac{-y}{x^2+y^2},\quad \frac{\partial u}{\partial y}=\pm\frac{x}{x^2+y^2}$$ and
so $$y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}= \pm\frac{-y^2-x^2}{x^2+y^2}=\mp 1.$$
Hence, we need $u(x,y)=-\arctan\frac{y}{x}$.
