$yU_x-xU_y=1, U(x,0)=0$ So I'm struggling a bit with solving this partial differential equation with this initial condition.
I'm familiar with the method of characteristics and have set up the equations as follows:
$$\frac{dx}{y}=\frac{dy}{-x}=du,$$
and taking the first and second expression and solving the ordinary differential equation assosciated to it gives me
$${c_{1}}=\frac{x^2+y^2}{2}.$$
Trying to do something similar for the first and third expressions gives me
$$\frac{dx}{y}=du,$$
or equivalently,
$$\frac{du}{dx}=\frac{1}{y},$$
and it's here that I am a bit stuck. Any help would be appreciated, thanks in advance.
EDIT: Thank you to everyone for the help.
 A: To solve $y \dfrac{\partial u}{\partial x} - x \dfrac{\partial u}{\partial y} = 1$:
Let $u = u(x(t),y(t))$.
Then, $\dfrac{\mathrm du}{\mathrm dt} = \dfrac{\partial u}{\partial x} \dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial u}{\partial y} \dfrac{\mathrm dy}{\mathrm dt}$.
Let:
$$\dfrac{\mathrm du}{\mathrm dt} = 1 \tag1$$
$$\dfrac{\mathrm dx}{\mathrm dt} = y \tag2$$
$$\dfrac{\mathrm dy}{\mathrm dt} = -x \tag3$$
Solving $(2)$ and $(3)$ gives $x^2+y^2=c$, whence $x=\pm\sqrt{c-y^2}$.
Dividing $(1)$ by $(3)$ gives, therefore:
$$\dfrac{\mathrm du}{\mathrm dy} = \dfrac1{\mp\sqrt{c-y^2}}$$
Which gives:
$$\int_0^u \mathrm du = \int_0^y \dfrac1{\mp\sqrt{c-y^2}}$$
(because $u=0$ when $y=0$)
Therefore, $u = \mp\arctan\left(\dfrac{y}{\sqrt{c-y^2}}\right) = -\arctan\left(\dfrac yx\right)$.

Checking:
We note that $\dfrac{\partial u}{\partial x} = \dfrac{y}{x^2+y^2}$ and $\dfrac{\partial u}{\partial y} = \dfrac{-x}{x^2+y^2}$.
Therefore, $y \dfrac{\partial u}{\partial x} - x \dfrac{\partial u}{\partial y} = \dfrac{y^2-(-x^2)}{x^2+y^2} = 1$.
Moreover, $u(x,0) = -\arctan\left(\dfrac0x\right) = 0$.
A: With characteristic equation
$$\frac{dx}{y}=\frac{dy}{-x}=du$$
you found
$${c_{1}}=\frac{x^2+y^2}{2}$$
and we have
$$\frac{ydx}{y^2}=\frac{xdy}{-x^2}=\frac{ydx-xdy}{y^2+x^2}=\frac{du}{1}$$
then $du=-\dfrac{xdy-ydx}{y^2+x^2}=-d\arctan\dfrac{y}{x}$ hence
$$u+\arctan\dfrac{y}{x}=c_2$$
finally
$$u=-\arctan\dfrac{y}{x}+f(\frac{x^2+y^2}{2})$$
A: $$\frac{dx}{y}=\frac{dy}{-x}=du$$
You correctly obtained a first characteristic equation :
$$x^2+y^2=C_1$$
This is equivalent to your equation with $C_1=2c_1$
$du=-\frac{dy}{x}=-\frac{dy}{\sqrt{C_1-y^2}} \quad\to\quad u=-\tan^{-1}\frac{y}{\sqrt{C_1-y^2}}+C_2$
$$ u+\tan^{-1}\frac{y}{x}=C_2$$
General solution : $\quad u+\tan^{-1}\frac{y}{x}=F(x^2+y^2)$
$$u(x,y)=-\tan^{-1}\frac{y}{x}+F\left(x^2+y^2\right)$$
The function $F(X)$ is to be determined according to the boundary condition :
$u(x,0)=0=-\tan^{-1}\frac{0}{x}+F\left(x^2\right)=F\left(x^2\right)=0\quad $any $x$.
This implies $F(X)=0$.
$$u(x,y)=-\tan^{-1}\frac{y}{x}$$
