Method of characteristics computation I am tryig to find all the solutions of the following equation:
$$(x+y)u_x+(u-x)u_y=(y+u)$$
I've started with the following system of equations:
$$\frac{dx}{x+y}=\frac{dy}{u-x}=\frac{du}{y+u}$$
to find a first and second integral, but i can't arrive to a relation that leads to a couple of ODE's, i've tryied sum's and substractions:
$$dx=\frac{x+y}{u-x}dy$$
$$du=\frac{u+y}{u-x}dy$$
$$dx-dy=\frac{(x+y)-(u+y)}{u-x}dy$$
$$dx-dy=\frac{x-u}{u-x}dy$$
$$dx-dy=-dy$$
$$dx+dy=du$$
Taking the integral at both sides:
$$x^2+y^2=u^2$$
$$u=\sqrt{x^2+y^2}$$
I don't know if this is an integral, or where to go from here, any insight will be greatly appreciated.
Kind Regards
 A: This is a fun case of first order quasilinear equations. Firstly, I definitely recommend checking out this post for a nice trick that will save you calculation time (Solving PDE $yu_x - xu_y + x^2 - y^2 = 0$ with method of characteristics). 
But by the very system of equations: 
$$\frac{dx}{x+y} = \frac{dy}{u-x} = \frac{du}{y+u}$$
I apply the same trick in the post I mentioned above: 
$$\frac{dx}{x+y} = \frac{dy}{u-x} \implies \frac{dx}{x+y} = \frac{dy}{u-x} = \frac{dx + dy}{y + u}$$
Now equating this with the last equation, I get: 
$$\frac{dx + dy}{y + u} = \frac{du}{y+u} \implies dx + dy = du$$
The only issue I see in your post is how you jump from 
$$dx + dy = du \overset{?}{\implies} x^2 + y^2 = z^2$$
If anything, this shouldn't be true, and instead we would conclude: 
$$z = x + y + C$$
Moreover, you can immediately check that in fact $z = x + y + C$ solves the equation. 
A: You made a mistake when you wrote :

Blockquote
  $$dx+dy=du$$
  Taking the integral at both sides:
  $$x^2+y^2=u^2$$

In fact integrating $dx+dy=du$ leads to 
$$u=x+y+\text{constant}$$
This is a particular solution of the PDE 
$$(x+y)u_x+(u-x)u_y=(y+u)$$
But this is far to be the general solution of the PDE. They are an infinity of solutions since no boundary condition is specified.
You correctly wrote the Charpit-Legendre system of ODEs :
$$\frac{dx}{x+y}=\frac{dy}{u-x}=\frac{du}{y+u}$$
A first characteristic equation is 
$$u-x-y=c_1$$
This can be found as you did (but without the mistake of integration, of course). 
A more direct method can be used, taking profit of a well-known property of the fractions : See the note at the end of my answer.
In fact, the particular solution $u=x+y+c_1$ can be seen without the above calculus, just by inspection of the PDE. 
A second characteristic equation comes from $\frac{dx}{x+y}=\frac{dy}{u-x}=\frac{dy}{(c_1+x+y)-x}=\frac{dy}{c_1+y}$
Solving $\frac{dx}{dy}=\frac{x+y}{c_1+y}$ leads to :
$x=c_1+(y+c_1)\ln|y+c_1|+c_2(y+c_1)$
$x=u-x-y+(u-x)\ln|u-x|+c_2(u-x)$
$$\frac{-u+2x+y-(u-x)\ln|u-x|}{u-x}=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is:
$$\boxed{\frac{-u+2x+y-(u-x)\ln|u-x|}{u-x}=F(u-x-y)}$$
$F$ is an arbitrary function, to be determined according to some boundary conditions (which are missing in the wording of the question).
NOTE :
If two fractions are equal :
$$\frac{A}{B}=\frac{C}{D}\:,$$
they are equal to any fractions on the form :
$$\frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}\:,$$
with any coefficients (constants or not) $c_1$ and $c_2$ not both nul.
For example in case of $\frac{dx}{x+y}=\frac{dy}{u-x}$
In your question : $A=dx\quad;\quad B=x+y\quad;\quad C=dy\quad;\quad D=u-x\quad;\quad c_1=c_2=1.$
$\frac{dx}{x+y}=\frac{dy}{u-x}=\frac{dx+dy}{(x+y)+(u-x)}=\frac{dx+dy}{y+u}$
Then $\frac{dx+dy}{y+u}=\frac{du}{y+u}\quad\implies\quad dx+dy=du$ .
