Question

 1) (y - xz)p + (yz - x)q = (x + y)(x - y)

Answer 1

Given equation is $(y - xz)p + (yz - x)q = (x + y)(x - y)$

Comparing with the standard equation, $Pp+Qq=R$

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

Equation will be,

$\frac{dx}{y-xz}=\frac{dy}{yz-x}=\frac{dz}{(x+y)(x-y)} = \frac{xdx+ydy+zdz}{0} = \frac{ydx+xdy}{y^2-x^2}$

$\frac{dx}{y-xz}=\frac{dy}{yz-x}=\frac{dz}{x^2-y^2} = \frac{xdx+ydy+zdz}{0} = \frac{ydx+xdy}{y^2-x^2}$

${xdx+ydy+zdz} = {0}$

solving it,

$x^2+y^2+z^2=c_1$ (1)

Equation (1) is the solution of the given differential equation

Taking (iii) and last part,

$\frac{dz}{x^2-y^2} = \frac{ydx+xdy}{y^2-x^2}$

$dz = -ydx-xdy$

Integrating both sides,

$z = -xy + c_2$  (2)

$z +xy = c_2$

Equations (1) and (2) are solutions to the equation.

The general solution of the differential equation will be

$\phi(x^2+y^2+z^2,z+xy)=0.$