Answer to Question #270361 in Differential Equations for Niru

Question

Find the integral surface of the equation x²p+y²q+z²=0 passing through z=1,x+y=xy

Answer 1

$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{-z^2}$

$-1/x=-1/y+c_1$

$1/z=-1/y+c_2$

$F(c_1,c_2)=F(1/y-1/x,1/z+1/y)=0$

for z=1,x+y=xy :

$c_2=1+1/y\implies y=c$

$c_1=1/y-1/x=\frac{x-y}{xy}=\frac{x-y}{x+y}=\frac{x-c}{x+c}$