Answer to Question #242424 in Differential Equations for Parag

Question

(x^2 -yz)p + (y^2 -zx)q = z^2 -xy

Answer 1

$(x^2-yz)p+(y^2-zx)q=z^2-xy\\ p=\frac{\delta z}{\delta x} \text{ and } q=\frac{\delta z}{\delta y}\\ \text{The auxilliary equations are }\\ \frac{dx}{x^2-yz}=\frac{dy}{y^2-zx}=\frac{dz}{z^2-xy}\\ \text{Hence }\\ \frac{dx-dy}{(x^2-yz)-(y^2-zx)}=\frac{dy-dz}{(y^2-zx)-(z^2-xy)}=\frac{dz-dx}{(z^2-xy)-(x^2-yz)}\\ \frac{d(x-y)}{(x-y)(x+y+z)}=\frac{d(y-z)}{(x-y)(x+y+z)}=\frac{d(x-y)}{(z-x)(x+y+z)}\\ \frac{d(x-y)}{x-y}=\frac{d(y-z)}{y-z}=\frac{d(z-x)}{z-x}\\ \text{Integrate both sides}\\ \ln|x-y|=\ln|y-z|+\ln C_1\\ \ln|y-z|=\ln|z-x|+ \ln C_2\\ \implies \frac{x-y}{y-z}=C_1\\ ~~~~~~~~~~\frac{y-z}{z-x}=C_2\\ \text{The general solution of the equation is }\\ \phi\left(\frac{x-y}{y-z},\frac{y-z}{z-x}\right)=0$