How to Separate Quasi-Linear PDE I'm attempting to solve the non-homogenous quasi-linear PDE below:
$$z\frac{\partial z}{\partial x} - \alpha y\frac{\partial z}{\partial y} = \frac{\beta}{x^3y}$$
From what I've read in texts, the general form of a quasi-linear PDE is defined as
$$a(x,y,z)\frac{\partial z}{\partial x} + b(x,y,z)\frac{\partial z}{\partial y} - c(x,y,z) = 0$$
with solutions (called characteristic curves) $\phi(x,y,z) = C_1$ and $\psi(x,y,z) = C_2$ given by the characteristic equations
$$\frac{dx}{a} = \frac{dy}{b} = \frac{dz}{c}$$
When I set up these equations for my problem, I find
$$a(x,y,z) = z$$
$$b(x,y,z) = -\alpha y$$
$$c(x,y,z) = \frac{\beta}{x^3 y}$$
which leads to
$$\frac{dx}{z} = -\frac{dy}{\alpha y} = \frac{x^3ydz}{\beta}$$
Due to the coupling in the last term I cannot find a way to separate these to get two expressions containing a total derivative.
Could anyone help?

In fact the main reason of why I get struck in How to Separate Charpit Equations is that when I directly apply the method of characteristics of the last equation, it involves a very difficult type of second-order nonlinear ODE:
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=-\alpha y$ , letting $y(0)=1$ , we have $y=e^{-\alpha t}$
$\begin{cases}\dfrac{dx}{dt}=z\\\dfrac{dz}{dt}=\dfrac{\beta}{x^3y}=\dfrac{\beta}{x^3e^{-\alpha t}}\end{cases}$
$\therefore\dfrac{d^2x}{dt^2}=\dfrac{\beta}{x^3e^{-\alpha t}}$
Which belongs to the ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0313.pdf
In fact the PDE is similar to the Burger's equation with source term.
