I have this equation $$ u_x+uu_y=0$$ by the book "Handbook of First order Partial Differential Equations - page 290." The general solution is $F(ux-y,u)$, where $F$ is a arbitrary function.
I try check this answer but I'm confuse. I start with this; the characteristic system is
$$\dfrac{dx}{1}=\dfrac{dy}{u}=\dfrac{u}{"0"},$$ so
$$\dfrac{du}{"0"} \rightarrow u=C_1$$ where $C_1$ is constant. And $$\dfrac{dy}{dx}=u \rightarrow ux-y = C_2$$ where $C_2$ is constant. Therefore $$C_1=F(C_2)\rightarrow F(ux-y)$$
I dont see how I have $F(C_1, C3)$ for exemple, how can I get $C_3$?
$$ u_x+uu_y=0$$
The first part is correct :
The characteristic system is
$$\dfrac{dx}{1}=\dfrac{dy}{u}=\dfrac{u}{"0"},$$ so
$$\dfrac{du}{"0"} \rightarrow u=C_1$$ where $C_1$ is constant. And $$\dfrac{dy}{dx}=u \rightarrow ux-y = C_2$$ where $C_2$ is constant.
At this point we already have 2 characteristics equations. There is no need for more : The implicit form is $F(C_1\, , \, C_2)$
Thus, the general solution expressed on implicit form is : $$F(u\, , \, ux-y)=0$$ where $F$ is any differentiable function of two variables.
This result can de presented on various equivalent implicit forms. For example : $$u=f(xu-y)$$ where $f$ is any differentiable function.
or $\quad u=\frac{1}{x}\left( y+g(u)\right) \quad$ where $g$ is any differentiable function.
But is is not possible to separate $u$ without loss of generality, in order to express $u(x,y)$ on explicit form.