Question

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

Answer 1

The equation

$$(x-y)p-(x-y+z)q=z$$

is a Lagrange linear PDE of the form:

$$Pp+Qq=R$$

where:

$P=x-y\\ Q=-(x-y+z)\\ R= z$

The auxiliary equation is of the form

$$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{z}$$

Thus, from the given equation, we have:

$$\frac{dx}{x-y}= \frac{dy}{-(x-y+z)}=\frac{dz}{z} \qquad \cdots (i)$$

Solving equation (i) we get:

$$\frac{dx+dy+dz}{x-y+[-(x-y+z)]+z} = k \qquad \text{(say)}\\ \frac{dx+dy+dz}{x-y-(x-y+z)+z} = k\\ \frac{dx+dy+dz}{x-y-x+y-z+z} = k\\ \frac{dx+dy+dz}{0} = k\\ \implies dx+dy+dz =0$$

Integrating through, we get:

$$\int dx+\int dy+\int dz = \int 0\\ x+y+z= c$$

this implies that:

$$\bold{c_1 = x+y+z}$$

From equation (i), we can have:

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

Integrating through, we get:

$$\int dx + \int dy = \int dz\\ x+y=z+c_2$$

This implies that:

$$\bold{c_2 = x+y-z}$$

The general solution is thus:

$$\phi(c_1,c_2) =0 \\\implies \phi(x+y+z\;,\quad x+y-z) =0$$