Question

Find the general solution of the Lagrange's equation 2yzp+zxq=3xy?

Answer 1

The auxiliary equations are

$\dfrac{dx}{2yz} = \dfrac{dy}{zx} = \dfrac{dz}{3xy}$

Taking the first two ratios,

$\begin{aligned} \dfrac{dx}{2yz} &= \dfrac{dy}{zx}\\ \dfrac{dx}{2y} &= \dfrac{dy}{x}\\ x~dx &= 2y~dy\\ \dfrac{x^2}{2} &= 2\dfrac{y^2}{2}+\dfrac{u}{2}\\ \dfrac{x^2}{2} - y^2 &= \dfrac{u}{2}\\ x^2 - 2y^2 &= u\\ \end{aligned}$

Taking the last two ratios,

$\begin{aligned} \dfrac{dy}{zx} &= \dfrac{dz}{3xy}\\ \dfrac{dy}{z} &= \dfrac{dz}{3y}\\ 3y~dy&= z~dz\\ \dfrac{3y^2}{2} &= \dfrac{z^2}{2} + \dfrac{v}{2}\\ 3y^2 -z^2&=v\\ \end{aligned}$

The general solution of the equation is,

$\begin{aligned} \phi(u,v)&=0\\ \phi(x^2-2y^2,3y^2-z^2)&=0 \end{aligned}$