# 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.

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. $$u(x,y)=\arctan\left(-\frac{y}{x}\right)$$
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 $u(x,y)=-\arctan\frac{y}{x}$.