first order partial differential equation I have a first order PDE:
$$xu_x+(x+y)u_y=1$$
With the initial condition:     
$$u(1,y)=y$$
I have calculated result in Mathematica:
$u(x,y)=\dfrac{y}{x}$ , but I am trying to solve the equation myself, but I had no luck so far. I tried with method of characteristics, but I could not get the correct results.
I would appreciate any help or maybe even whole procedure.  
Thanks
 A: There must be (at least) an error here, but I fail to see it: 
General first order two variables PDE:
$$a(x,y)u_{x}(x,y)+b(x,y)u_{y}(x,y)+c(x,y)u(x,y)=f(x,y) $$
In this case
$a(x,y)=x$, $b(x,y)=(x+y)$, $c(x,y)=0$ and $f(x,y)=1$
Now we will find some $$\xi(x,y)=c $$
$c \in \mathbb{R}$ constant solves the differential equation
$$\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{x+y}{x}=1+\frac{y}{x} $$
In order to solve this define a new function $$u(x)=\frac{y(x)}{x}$$ so that
$$\frac{du}{dx}=\frac{dy}{dx}\frac{1}{x}-\frac{y(x)}{x^2}=\left(1+u\right)\frac{1}{x}-\frac{u}{x}=1 $$
Hence $u(x)=x+c$, $\frac{y}{x}=x+c \rightarrow c=\frac{y}{x}-x$
$$\boxed{\xi(x,y)=\frac{y}{x}-x} \Rightarrow (\xi +x)x=y$$
Now let $u(x,y)=v(x,\xi)$ and take derivatives
$$u_{x}=v_{x}+v_{\xi}\left( -\frac{y}{x^2}-1\right) =v_{x}+\left( -\frac{(\xi+x)}{x}-1\right)v_{\xi}$$
$$ u_{y}=v_{\xi}\frac{1}{x}$$
Substitution into the PDE yields
$$ x\left( v_{x}+\left( -\frac{(\xi+x)}{x}-1\right)v_{\xi}\right)+(x+(\xi+x)x)v_{\xi}\frac{1}{x}=1$$
$$xv_{x}-(\xi+x)v_{\xi} -xv_{\xi}+v_{\xi}+\xi v_{\xi}+xv_{\xi}=1$$
$$v_{x}x+v_{\xi}(1-x)=1$$
A: $xu_x+(x+y)u_y=1$
$u_x+\left(\dfrac{y}{x}+1\right)u_y=\dfrac{1}{x}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(1)=1$ , we have $x=t$
$\dfrac{dy}{dt}=\dfrac{y}{x}+1=\dfrac{y}{t}+1$ , letting $y(1)=y_0$ , we have $y=y_0t+t\ln t=y_0x+x\ln x$
$\dfrac{du}{dt}=\dfrac{1}{x}=\dfrac{1}{t}$ , letting $u(1)=f(y_0)$ , we have $u(x,y)=\ln t+f(y_0)=\ln x+f\left(\dfrac{y}{x}-\ln x\right)$
$u(1,y)=y$ :
$f(y)=y$
$\therefore u(x,y)=\ln x+\dfrac{y}{x}-\ln x=\dfrac{y}{x}$
A: my answer   $\frac{dx}{x}=\frac{dy}{x+y}=\frac{du}{1}$
from first and third eqn.  $\frac{exp(u)}{x}=c_1$
from first and second eqn
$\frac{dy}{dx} -\frac{y}{x}=1$ which is first order ode and it can be easily solved  $\frac{y}{x}- ln x= c_2$
 parametrized initial curve C : x=1 , y=t , u=t
$u_1 =\frac{exp(u)}{x}$ and $u_2= \frac{y}{x}- ln x$ are two first integrals
on initial curve say u_1|C =U_1= exp t, u_2|C =U_2= t eliminating t we find U_1=exp U_2
come back U_1=u_1, U_2=u_1 we obtain u= y/x
