Answer to Question #209167 in Differential Equations for Abbot

Question

xzp - yzq= Kxyz by lagrange method

Answer 1

This is Lagrange's equation

$$Pp+Qq=R$$

$$P=xz, Q=-yz, R=Kxyz$$

The auxiliary equation is,

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

$$\dfrac{dx}{xz}=\dfrac{dy}{-yz}=\dfrac{dz}{Kxyz}$$

$$\dfrac{dx}{xz}=\dfrac{dy}{-yz}=>\dfrac{dx}{x}=-\dfrac{dy}{y}$$

$$\int \dfrac{dx}{x}=-\int\dfrac{dy}{y}$$

$$\ln x=-\ln y+\ln C_1$$

$$xy=C_1$$

$$\dfrac{dx}{xz}=\dfrac{dz}{Kxyz}$$

$$\dfrac{dx}{xz}=\dfrac{dz}{Kx(\dfrac{C_1}{x})z}$$

$$KC_1\dfrac{dx}{x}=dz$$

$$KC_1\ln x=z+C_2$$

$$Kxy\ln x-z=C_2$$

Hence, the required general solution is given by

$$\phi(xy, Kxy\ln x-z)=0$$

where $\phi$ is an arbitrary function.