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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: $u^2−y^2=f(x^2−y^2)$

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    Your English is just fine, but the question would gain in clarity if you used formulae or even an example. Prose hasn't been used in mathematics for about two centuries, anymore, precisely because mathematics done in prose is ambiguous. – 2017-01-05
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    I add a comment instead of editing my question. – 2017-01-05
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    On the contrary, editing your question to add more details is encouraged and recommended. – 2017-01-05
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    I add a comment instead of editing my question. Consider this PDE $yuu_x + xuu_y = xy$ Solving the ODE's we get first $x^2 - y^2 = c_1$ and $u^2 - y^2 = c_2$. Now we make this identification (the one I don't understand where does it come from) $c_1 = f(c_2)$ being f(dummy) an arbitrary function. We obtain the general solution: $u^2 - y^2 = f(x^2 - y^2)$ – 2017-01-05
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    Too late your comment :) – 2017-01-05

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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.