Quasi-linear PDE $xu_x+yu_y=1$ with $u=y$ over unit circle I am given the following equation.

$xu_x+yu_y=1$ with the conditions $u(x,y)=y$ for all $x^2+y^2=1$

What I got so far.
I calculated the characteristics $x'(t)=x(t),x(0)=x_0\in S_1$ which are $x(t)=\exp(t)*xo$
$U(t)=u(x(t))$
Now I need to solve the equation $U' =1$ $\implies$ $U(t)=t+w_2$ where $w_2$ is the second component of $x_0$
How can I derive the solution $u(x,y)$ from that ?
 A: Provided that, for any $t,s\in\mathbb{R}$,
$$
\frac{\rm d}{{\rm d}t}u(e^{t+s},e^t)=e^{t+s}\frac{\partial u}{\partial x}(e^{t+s},e^t)+e^t\frac{\partial u}{\partial y}(e^{t+s},e^t)=\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)(e^{t+s},e^t)=1,
$$
we have
$$
u(e^{t+s},e^t)=u(e^{t_0+s},e^{t_0})+\int_{t0}^t1{\rm d}t=u(e^{t_0+s},e^{t_0})+t-t_0.
$$
Further, when $t_0<0$, set $s=-t_0+\log\sqrt{1-e^{2t_0}}\in\mathbb{R}$. This yields, as per the given boundary condition,
$$
u(e^{t-t_0+\log\sqrt{1-e^{2t_0}}},e^t)=u(\sqrt{1-e^{2t_0}},e^{t_0})+t-t_0=e^{t_0}+t-t_0.
$$
Finally, thanks to the arbitrariness of $t$, set
\begin{align}
x&=e^{t-t_0+\log\sqrt{1-e^{2t_0}}}>0,\\
y&=e^t>0,
\end{align}
whose inverse reads
\begin{align}
t&=\log y\in\mathbb{R},\\
t_0&=-\frac{1}{2}\log\left[1+\left(\frac{x}{y}\right)^2\right]<0.
\end{align}
Thus when $x,y>0$, we have
$$
u(x,y)=\left[1+\left(\frac{x}{y}\right)^2\right]^{-1/2}+\log y+\frac{1}{2}\log\left[1+\left(\frac{x}{y}\right)^2\right]=\frac{y}{\sqrt{x^2+y^2}}+\log\sqrt{x^2+y^2}.
$$
Note that the above expression satisfies the boundary condition for not only the $x,y>0$ case, but all cases. Therefore, we conclude that
$$
u(x,y)=\frac{y}{\sqrt{x^2+y^2}}+\log\sqrt{x^2+y^2}
$$
for all $\left(x,y\right)\in\mathbb{R}^2\setminus\left\{\left(0,0\right)\right\}$.
A: Considering the change of variables
$$
\begin{array}{rcl}
x & = & r\cos\theta\\
y & = & r\sin\theta
\end{array}
$$ 
with 
$$
\begin{array}{rcl}
dx & = & dr\cos\theta-r\sin\theta d\theta\\
dy & = & dr\sin\theta+r\cos\theta d\theta
\end{array}
$$
we have
$$
\begin{array}{rcl}
u_{x} & = & u_{r}\frac{dr}{dx}+u_{\theta}\frac{d\theta}{dx}\\
u_{y} & = & u_{r}\frac{dr}{dy}+u_{\theta}\frac{d\theta}{dy}
\end{array}
$$
Here $\frac{dx}{x}=\frac{dy}{y}\Rightarrow\frac{dy}{dx}=\frac{y}{x}=\tan\theta$
so we obtain
$$
xu_{x}+yu_{y}=1\Longleftrightarrow2ru_{r}=1\Rightarrow u(r,\theta)=\ln(\sqrt{r})+\Phi(\theta)
$$
Now with the boundary conditions
$$
u(1,\theta)=\sin\theta\Rightarrow\Phi(\theta)=\sin\theta
$$
and finally
$$
u(r,\theta)=\ln(\sqrt{r})+\sin\theta
$$
or in $(x,y)$ coordinates, 
$$
u(x,y)=\frac{1}{2}\ln(x^{2}+y^{2})+\frac{y}{\sqrt{x^{2}+y^{2}}}
$$
