I have been reading some notes about Quasi Linear PDE and I stuck to the following issue:
Consider the PDE: $xz_x+yz_y=xe^{-z}$
then my notes read as:
"the general solution is: $G(x/y, e^z-x)=0$ . (1)
and so $$e^z-x=g(x/y)$$ , where $G,g$ are $C^1$ functions."
In my opinion this is not right and we are losing solutions that way.
REMARK: The only way I know we can do a similar thing is when the following PDE
$az_x+bz_y=0$ (2) is given, where $a,b$ are functions of $x,y$ and they do not vanish simultaneously. Then if $u=c$ is the general solution of the ODE:
$$\frac{dx}{a}=\frac{dy}{b}$$, then $z=f(u)$ , where $f$ is any $C^1$ function.
This last is taken from the book Introduction to PDE with applications by Zachamanoglou and Thoe, Example 2.2 page 62.
Is my cosideration correct?
$$xu_x+yu_y=xe^{-u}$$ The Charpit-Lagrange equations are : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{xe^{-u}}$$ First equation of characteristic curves , from $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{x}{y}=c_1$$ Second equation of characteristic curves , from $\frac{dx}{x}=\frac{du}{xe^{-u}}$: $$e^u-x=c_2$$ The general solution of the PDE, expressed on the form of implicit equation is : $$\Phi\left(\frac{x}{y}\:,\:e^u-x\right)=0$$ where $\Phi$ is an arbitrary function of two variables.
Or, equivalently on explicit form : $e^u-x=F\left(\frac{x}{y}\right)$ $$u(x,y)=\ln\left|x+F\left(\frac{x}{y}\right)\right|$$ where $F$ is an arbitrary function.
The function $F$ had to be determined according to some boundary condition. Since there is no such condition specified in the wording of the problem, further calculus is not possible.
Nevertheless, the above result is consistent with the solution (which is correct) noted in the OP question.