Elimination of constants in pde's. How does it appears the arbitrary function in the general solution? Considering the method of characteristics to solve quasi-linear PDE's, I don't totally catch why one of the constants from the ODE's are related functionally one to the other (two independent variables only, if necessary to specify). I understand how it works and how the arbitrary function as part of the general solution appears, but not why the constants have such a relation.
Somehow, and here is where I need some clarification, the point is that the two families of curves must be in the same surface solution. I cannot "jump" from here to the functional dependence of the constants.
Sorry for my english and thanks in advance.
EDIT ADDED:
(Without errors, I hope)
Consider this PDE $yuu_x+xuu_y=xy$ Solving the associated ODE's we get first $x^2−y^2=c_1$ and second $u^2−y^2=c_2$. Now we make this identification (the one I don't understand where does it come from) $c_2=f(c_1)$ being f a single variable arbitrary function. We obtain the general solution: $u^2−y^2=f(x^2−y^2)$ 
 A: It is doubtful that one could explain in a few words what requires several pages in books where the theory of the method of characteristics is presented.
For example : http://www.ehu.eus/luis.escauriaza/apuntes_problemas_y_examene/method-of-characteristics.pdf
In the case of the PDE $\quad yuu_x+xuu_y=xy\quad$ the set of differential characteristic equations can be presented on various manner. A summary :
$$\frac{dx}{uy}=\frac{dy}{ux}=\frac{du}{xy}$$
A first family of characteristic curves comes from : $\frac{dx}{uy}=\frac{dy}{ux} \quad\to\quad xdx-ydy=0 \quad\to\quad x^2-y^2=c_1$
A second family of characteristic curves comes from : $\frac{dy}{ux}=\frac{du}{xy} \quad\to\quad udu-ydy=0 \quad\to\quad u^2-y^2=c_2$
One can even find a third family of characteristic curves coming from : $\frac{dx}{uy}=\frac{du}{xy} \quad\to\quad udu-xdx=0 \quad\to\quad u^2-x^2=c_3 \quad$ 
But whey are not independent since $c_3=c_2-c_1$. Doesn't matter the couple of them that we chose among the three.
For example, with the first and the second, all above is valid with independent $c_1$ and $c_2$ on the respective characteristic curves. This is no longer true on the surface linking two independent characteristic curves : To represent a solution of the PDE, $c_1$ and $c_2$ are no longer independent. The relationship can be expressed on the form of $\quad \Phi(c_1\:,\:c_2)=0\quad$ , where $\Phi$ is any differentiable function of two variables. Thus, an implicit form of the general solution of the PDE is :
$$\Phi(x^2-y^2\:,\:u^2-y^2)=0$$
As well, another equivalent form is $\quad \phi(x^2-y^2\:,\:u^2-x^2)=0\quad$ where $\phi$ and $\Phi$ are any functions of two variables, but related functions.
Other equivalent relationships between $c_1$ and $c_2$ is to express one of then as a function of the other : $\quad c_1=F(c_2)\quad$ or $\quad c_2=f(c_1)\quad$ with any differentiable functions $F$ and $f$, but related functions. Thus, alternative equivalent forms for the general solution are :
$$x^2-y^2=F(u^2-y^2) \quad \text{or} \quad u^2-y^2=f(x^2-y^2)$$
Boundary conditions are necessary to determine those functions.
