Similar to Method of characteristics inhomogeneous nonlinear wave equation:
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$
$\begin{cases}\dfrac{dx}{dt}=u\\\dfrac{du}{dt}=x\end{cases}$
$\therefore\dfrac{d^2x}{dt^2}=x$
$x=C_1\sinh t+C_2\cosh t$
$\therefore u=C_1\cosh t+C_2\sinh t$
Hence $\begin{cases}x=C_1\sinh t+C_2\cosh t\\u=C_1\cosh t+C_2\sinh t\end{cases}$
$x(0)=x_0$ , $u(0)=F(x_0)$ :
$\begin{cases}C_1=F(x_0)\\C_2=x_0\end{cases}$
$\therefore\begin{cases}x=F(x_0)\sinh t+x_0\cosh t\\u=F(x_0)\cosh t+x_0\sinh t\end{cases}$
$\therefore\begin{cases}x_0=x\cosh t-u\sinh t=x\cosh\ln y-u\sinh\ln y\\F(x_0)=u\cosh t-x\sinh t=u\cosh\ln y-x\sinh\ln y\end{cases}$
Hence $u\cosh\ln y-x\sinh\ln y=F(x\cosh\ln y-u\sinh\ln y)$
$u(x,1)=2x$ :
$F(x)=2x$
$\therefore u\cosh\ln y-x\sinh\ln y=2x\cosh\ln y-2u\sinh\ln y$
$u(x,y)=\dfrac{x(\sinh\ln y+2\cosh\ln y)}{2\sinh\ln y+\cosh\ln y}$
$$\begin{cases}
xu_x+(x+y)u_y=1\\
u(1,y)=y\\
\end{cases}
$$
$$\frac {dx}{x}=\dfrac {dy}{x+y}=\dfrac {dz}{1}$$
So you need to solve this system of DE:
$$\begin{cases}\dfrac {dx}{x}=\dfrac {dy}{x+y} \\ \dfrac {dz}{1}=\dfrac {dx}{x} \tag{2}\end{cases}$$
$$\dfrac {dx}{x}=\dfrac {dy}{x+y} $$
$$(x+y)dx=xdy $$
$$ydx-xdy=-xdx$$
$$\frac {ydx-xdy}{x^2}=-\frac 1 x dx$$
$$\frac {xdy-ydx}{x^2}=\frac 1 x dx$$
$$d\left (\frac y x \right )= \frac 1 x dx$$
Integrate:
$$ y (x) = x\ln x+C_2x$$
The second one is easy to integrate:
$$\dfrac {dz}{1}=\dfrac {dx}{x} \tag{2}$$
$$z+C_1=\ln x$$
You can surely take from there.
Best Answer
$$xu_x+(x+y)u_y=1$$ You have got the correct system of equation which can be written on this form : $$\frac{dx}{x}=\frac{dy}{x+y}=\frac{du}{1}=dt$$ Solving $\frac{dx}{x}=\frac{dy}{x+y}$ gives a first characteristic $$\frac{y}{x}-\ln|x|=c_1$$ Solving $\frac{dx}{x}=\frac{du}{1}$ gives a second characteristic $$u-\ln|x|=c_2$$ The general solution of the PDE is $u-\ln|x|=F\left(\frac{y}{x}-\ln|x|\right)$ $$u(x,y)=\ln|x|+F\left(\frac{y}{x}-\ln|x|\right)$$ where $F$ is an arbitrary function.
This function is determined according to the boundary condition :
$u(1,y)=y=\ln|1|+F\left(\frac{y}{1}-\ln|1|\right)=F(y)$
Thus $F(y)=y$ and as a consequence $F\left(\frac{y}{x}-\ln|x|\right)=\frac{y}{x}-\ln|x|$ .
$u(x,y)=\ln|x|+\left(\frac{y}{x}-\ln|x|\right)$
The final solution is : $$u(x,y)=\frac{y}{x}$$