Solving $(y+u) u_x + y u_y = x - y $ I have the following problem:
$$(y+u) u_x + y u_y = x - y $$
I only know characteristic curves so I did:
$\frac{dx}{dt} = y + u $
$\frac{dy}{dt} = y$
I began solving the second equation resulting $y = C_1 e^t$. Using that result I solved  $\frac{dx}{dt} = C_1 e^t + u  $. This is where my doubts began, since $x$ has $u$. I decided to give it a try so I got $x = C_1 e^t + u t $. Here there's a constant that I've been said the method omits. And I've seen some examples where it's omitted. So I continued by using the chain rule, ending up in this equation:
$$\frac{du}{dt} = x - y = (C_1 e^t + u t) - C_1 e^t = u t $$
Solving this gives $u = e^{\frac{t^2}{2}} + K $ where I know $K$ is an arbitrary function. Replacing $t$ from $y = C_1 e^t$ we get $$u = \ln \big(\frac{y}{C_1}\big) + K $$ but clearly this is incomplete. I don't know the form of $K(x,y)$ and $C_1$. I would like to know how to manage the arbitrary functions since I feel it's what's happening, and how to manage them in this case where we have a quasilinear PDE. Also, if there's another way to solve this, I don't know if substracting $\frac{dx}{dt}$ and $\frac{dy}{dt}$ helps in some way since it leaves $u$ or if there's a common change of variables that's useful in these problems. Thanks for the help.
 A: $$(y+u) u_x + y u_y = x - y $$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{y+u}=\frac{dy}{y}=\frac{du}{x-y}$$
FIRST :
$\frac{dx}{y+u}=\frac{du}{x-y}=\frac{dx+du}{(y+u)+(x-y)}=\frac{dx+du}{x+u}$
A first characteristic equation comes from solving $\quad \frac{dy}{y}=\frac{dx+du}{x+u}$
$$\frac{u+x}{y}=c_1$$
SECOND :
$\frac{dy}{y}=\frac{du}{x-y}=\frac{dy+du}{(y)+(x-y)}=\frac{dy+du}{x}$
A second characteristic equation comes from solving $\quad \frac{dx}{y+u}=\frac{dy+du}{x}$
$$(y+u)^2-x^2=c_2$$
GENERAL SOLUTION :
The general solution of the PDE can be expessed on the form of various equivalent implicit equations , for example :
$$\Phi\left(\frac{u+x}{y}\,,\,\left((y+u)^2-x^2\right)\right)=0$$
where $\Phi$ is an arbitrary function of two variables.
Or :
$$\frac{u+x}{y}=F\big((y+u)^2-x^2\big)$$
where $F$ is an arbitrary function.
Or :
$$(y+u)^2-x^2=G\big(\frac{u+x}{y}\big)$$
where $G$ is an arbitrary function.
Other equivalent implicit equations can be presented with other arbitrary functions involving $c_1$ and $c_2$ , all related one to another.
Some boundary condition has to be specified to determine one of those functions and thus the solution satisfying both the PDE and the condition.
NOTE :
If you prefer the method with parameter you can write
$$\frac{dx}{y+u}=\frac{dy}{y}=\frac{du}{x-y}=dt$$
Separate the three ODEs and combine them in order to solve for functions of the parameter $t$.
