Question

(3x+y-z)p + (x+y-z)q = 2(z-y)

Answer 1

We have a quasilinear first-order PDE.

$p=\frac{\partial z}{\partial x},\ q=\frac{\partial z}{\partial y}.$

Characteristic system of the equation:

$\frac{dx}{3x+y-z}=\frac{dy}{x+y-z}=\frac{dz}{2(z-y)}$

We will find two independent particular solutions of this system.

Each ratio is equal to $\frac{-dx+3dy+dz}{0}.$ So $-dx+3dy+dz=0.$ Then $x-3y-z=C_1\text{ is the first solution of the system}.$

Each ratio is equal to $\frac{dx-dy+dz}{2x-2y+2z}=\frac{dx+dy-dz}{4x+4y-4z}.$

$\frac{d(x-y+z)}{x-y+z}=\frac{d(x+y-z)}{2(x+y-z)}.$

On integrating both sides we get:

$\ln (x-y+z)=\frac{1}{2}\ln (x+y-z)+\ln C_2.\\ \frac{x-y+z}{\sqrt{x+y-z}}=C_2\text{ is the second solution of the system}.$

Then the general solution of our equation can be written as

$\Phi(x-3y-z; \frac{x-y+z}{\sqrt{x+y-z}})=0$

where $\Phi$ is an arbitrary function of two variables.