Solve for $u$ the PDE $(x − y − 1)u_x + (y − x − u + 1)u_y = u$ if $u=1$ on $x^2+(y+1)^2=1.$ 
Solve the Cauchy problem $$(x − y − 1)u_x + (y − x − u + 1)u_y = u,$$
  if $u=1$ on $x^2+(y+1)^2=1.$

Attempt. $$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$
so $$\frac{dx+dy}{(x-y-1)+(y-x-z+1)}=\frac{dz}{z}\iff d(x+y+z)=0$$
so $g_1(x,y,z)=x+y+z=c_1.$ I have not managed to find the second relation $g_2(x,y,z)=0$, needed in order to get $F(g_1,g_2)=0$ for some $F$. (As far as I am concerned, there are no standard procedures in these cases, one has to work on trial-and -error to find the exact expressions).
Thanks in advance.
 A: For the second constant of integration
$$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$
$$\frac{d(x+y)}{x+y-k}=\frac{dy}{2y-k+1}$$
$$z^2=C({2y + {1-k}})$$
Where $k=x+y+z$
$$z^2=C({y-x-z + 1})$$
A: $$(x − y − 1)u_x + (y − x − u + 1)u_y = u \tag 1$$
$$\frac{dx}{x-y-1}=\frac{dy}{y-x-u+1}=\frac{du}{u}\quad\text{is correct}$$
You rightly found a first characteristic equation :
$$x+y+u=c_1$$
A second characteristic equation comes from 
$$\frac{dx-dy}{(x-y-1)-(y-x-u+1)}=\frac{du}{u}=\frac{d(x-y-1)}{2(x-y-1)+u}$$
With $v=x-y-1$
$$\frac{du}{u}=\frac{dv}{2v+u}$$
is a separable ODE which solution is $v=-u+c_2u^2$ . Thus $x-y-1=-u+c_2u^2$ . The second characteristic equation is :
$$\frac{x-y-1+u}{u^2}=c_2$$
The general solution of the PDE can be expressed on the form of implicit equation :
$$\frac{x-y-1+u}{u^2}=F(x+y+u) \tag 2$$
where $F$ is an arbitrary function.
Boundary condition in order to determine the function $F$ :
$u=1$ on $x^2+(y+1)^2=1$ , so $\frac{x-y-1+1}{1^2}=F(x+y+1)$
$$F(x+y+1)=x-y$$
Let $X=x+y+1=\pm\sqrt{1-(y+1)^2}+y+1$
$(X-y-1)^2=1-(y+1)^2.\quad$ To be solved for $y$ which leads to 
$y=\frac{X-2\pm\sqrt{2-X^2}}{2}\quad;\quad x=\frac{X\mp\sqrt{2-X^2}}{2}\quad;\quad x-y=\mp\sqrt{2-X^2}+1$
$$F(X)=1\mp\sqrt{2-X^2}$$
So, $F(X)$ is determined. We put it into the above general solution $(2)$ where $X=x+y+u$.
$$\frac{x-y-1+u}{u^2}=1\mp\sqrt{2-(x+y+u)^2}$$
The solution which complies to the PDE and the boundary condition is expressed on the form of an implicit equation :
$$x-y-1+u-u^2\pm u^2\sqrt{2-(x+y+u)^2}=0 \tag 3$$
To express $u(x,y)$ on explicit form we have to solve a polynomial equation of sixth degree. Thus there is no closed form for $u(x,y)$. The final answer is the above implicit form $(3)$.
