Answer to Question #289453 in Differential Equations for Pankaj

Question

Find the integral surface of the equation:

(x²-yz)p+(y²-zx)q =z²-xy

passing through the line x=1 , y=0

Answer 1

$\text{To find the integral surface of the given equation, we proceed as follows:}$

The auxiliary equations are;

$\displaystyle \frac{dx}{x^2-yz}=\frac{dy}{y^2-zx}=\frac{dz}{z^2-xy}\\$

$\displaystyle \Rightarrow \frac{dx-dy}{x^2-yz-y^2+zx}=\frac{dy-dz}{y^2-zx-z^2+xy}=\frac{dz-dx}{z^2-xy-x^2+yz}\\ \Rightarrow \frac{dx-dy}{(x-y)(x+y+z)}=\frac{dy-dz}{(x+y+z)(y-z)}=\frac{dz-dx}{(x+y+z)(z-x)}\\ \Rightarrow \frac{dx-dy}{x-y}=\frac{dy-dz}{y-z}=\frac{dz-dx}{z-x}\cdots\cdots\cdots\cdots(1)\\$

Integrating first two members of (1) yields;

$\displaystyle \log(x-y)=\log(y-z)+\log c_1\\ \Rightarrow\log \frac{x-y}{y-z}=\log c_1\\ \Rightarrow \frac{x-y}{y-z}=c_1\\$

Similarly from the last two members of (1), we have;

$\displaystyle \frac{y-z}{z-x}=c_2$

The required solution is;

$\displaystyle f\left[\frac{x-y}{y-z},\frac{y-z}{z-x}\right]=0$