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.
 A: You have done well so far. The system of ODE's is correct. Notice that $\xi=0$ due to your condition $u(x,0)=0$. However, your method to invert the $x$ and $y$ coordinates does not lead you anywhere. The following method to solve the system will be more useful.
$$\frac{d}{dt}\left(\frac{dx}{dt}=y\right)\implies\frac{d^2x}{dt^2}=\frac{dy}{dt}=-x$$
$$\implies x=c_1\sin(t)+c_2\cos(t)$$
Now we can find expression for $y$.
$$\frac{dy}{dt}=-c_1\sin(t)-c_2\cos(t)\implies y=c_1\cos(t)-c_2\sin(t)$$
We now use the given boundary condition $u(x,0)=0$ to find the constants $c_1$ and $c_2$.
$$y(0)=c_1\implies c_1=0$$
$$x(0)=c_2\implies c_2=x_0$$
Thus
$$x=x_0\cos(t)\quad\&\quad y=-x_0\sin(t)$$
We must invert these equations. Because they are nonlinear, we must be clever. First divide $y$ by $x$.
$$\frac{y}{x}=-\frac{\sin(t)}{\cos(t)}=-\tan(t)$$
$$\implies t=\arctan\left(-\frac{y}{x}\right)$$
Since you found $u=t$, we do not need to find $x_0$ and have arrived at our solution.
$$u(x,y)=\arctan\left(-\frac{y}{x}\right)$$
A: 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 $u(x,y)=-\arctan\frac{y}{x}$.
