Find the solution for the equation $(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y} = x+y$ Solve the equation $(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y} = x+y$ where $u=u(x,y)$. 
What I did:
Let v(x, y, u) = 0 and by changing u into v the PDE becomes:
$(y+z) \frac{\partial v}{\partial x} + (x+z) \frac{\partial v}{\partial y} + (x+y)\frac{\partial v}{\partial u}=0$
To solve the equation I have to find the solutions for the system
$\frac{dx}{y+z}=\frac{dy}{x+z}=\frac{du}{x+y}$
Here I got stuck. I know how to solve it when u=z, but since in this case I know nothing about z, I'm not sure how to proceed.
 A: The question is : Solve the equation $(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y} = x+y$ where $u=u(x,y)$. 
This is contradictory because $z$ appears in the PDE but it doesn't appears in $u(x,y)$.
They are two different ways to understand the problem :
FIRST, the variable $z$ was forgotten in the function $u$ :
The question should be : Solve the equation $(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y} = x+y$ where $u=u(x,y,z)$. 
$$(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y}+0*\frac{\partial u}{\partial z}=x+y$$
So, the Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dx}{y+z}=\frac{dy}{x+z}=\frac{dz}{0}=\frac{du}{x+y}$$
Thus one characteristic equation is 
$$z=c_1$$
and two other characteristic equations will come from :
$$\frac{dx}{y+c_1}=\frac{dy}{x+c_1}=\frac{du}{x+y}$$
I suppose that you can solve them and find the general solution $u(x,y,z)$ which includes an arbitrary function.
SECOND, the writting of the PDE equation is ambiguous because two different symbols are used for a same function :
$$u(x,y)=z(x,y)$$
The question should be : Solve the equation 
$$(y+u) \frac{\partial u}{\partial x} + (x+u) \frac{\partial u}{\partial y} = x+y \text{ where } u=u(x,y)$$
So, the Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dx}{y+u}=\frac{dy}{x+u}=\frac{du}{x+y}$$
I suppose that you can solve them for two characteristic equations and find the general solution $u(x,y)=z(x,y)$ which includes an arbitrary function.
To avoid the ambiguity you should write down which of the two above problems is the correct one.
If you are stuck in the calculus, you are wellcome to show where is the difficulty.
A: Equation: $(y+z) \frac{\partial u}{\partial x} + (x+z) \frac{\partial u}{\partial y} = x+y$. We regard z as constant. 
Let us define $x' \equiv x+z$ , $y' \equiv y + z$, and $\,\, v(x',y') \equiv u(x',y')-x'-y'$.
Then Equation: $y' \frac{\partial v}{\partial x'} + x' \frac{\partial v}{\partial y'} = -2z$.
In fact, we have some strategy for solving above pde.

For now, I regard x', y' as x , y.
Consider transformation $(s,t) = (\frac{y^2 - x^2}{2},x+y)$. Then relation between partial diffrential is 
$$
\begin{bmatrix} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial s}{\partial x} &\frac{\partial t}{\partial x} \\ \frac{\partial s}{\partial y} & \frac{\partial t}{\partial y}\end{bmatrix} \begin{bmatrix} \frac{\partial}{\partial s}\\ \frac{\partial}{\partial t} \end{bmatrix} =\begin{bmatrix} -x &1 \\ y & 1\end{bmatrix} \begin{bmatrix} \frac{\partial}{\partial s}\\ \frac{\partial}{\partial t} \end{bmatrix}
$$
Then equation will be 
$$
y \frac{\partial v}{\partial x} + x \frac{\partial v}{\partial y} =y(-x\frac{\partial v}{\partial s} + \frac{\partial v}{\partial t}) + x(y\frac{\partial v}{\partial s} + \frac{\partial v}{\partial t}) = \pmb{t \frac{\partial v}{\partial t} = {-2z}}
$$
i.e Solution:
$$
v = -2zln(|t|) + F(s)\quad .\\
$$

Equivalently,
$$
v = -2z\,ln(|x'+y'|) + F({y'}^2 - {x'}^2)\\
u = -2z\,ln(|x'+y'|) + F({y'}^2 - {x'}^2) + x' + y'
$$
Thus,
$$
u = -2z\,ln(|x+y+2z|) + x + y + 2z + F[\,(x-y)(x+y+2z)\,]
$$
$$$$
Any question or indication will be appreciated.
