Question

Find general solution of x(z^2-y^2)p+y(x^2-z^2)q=z(y^2-x^2)

Answer 1

We have to solve the following differential equation given by $x(z^2-y^2)p+y(x^2-z^2)q=z(y^2-x^2)$

Solution:-

$x(z^2-y^2)p+y(x^2-z^2)q=z(y^2-x^2)......$(1)

it is Lagrange's partial differential equation of type

$Pp+Qq=R$ ;

So, it has the solution which is

Lagrange's auxiliary equation such that

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$$.......(2)$

So,

the auxiliary equation of the equation (1) is

$\frac{dx}{x(z^2-y^2)}=\frac{dy}{y(x^2-z^2)}=\frac{dz}{z(y^2-x^2)}\\=\frac{xdx}{x^2(z^2-y^2)}=\frac{ydy}{y^2(x^2- z^2)}=\frac{zdz}{z^2(y^2-x^2)}=\frac{xdx+ydy+zdz}{0}\\\implies xdx+ydy+zdz=0\\\implies \int xdx+\int ydy+\int zdz=0\\\implies x^2+y^2+z^2=c.........(3)$

equation (3) is the general solution of the given equation .