Answer to Question #152921 in Differential Equations for meron

Question

solve (xz-y)p + (yz - x)q =1-z^2

Answer 1

P=zx-y

Q=zy-x

R=1-z²

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

$\frac{dx}{zx-y}=\frac{dy}{zy-x}=\frac{dz}{1-z^2}$

Multiplyers: z,1,x

zdx+dy-xdz=0

$zx+y+zx=2zx+y=C_1$

Multiplyers: 1,z,y

dx+zdy+ydz=0

$x+zy+zy=x+2zy=C_2$

The general solution is,

$\phi(c_{1},c_{2})=0\\ \phi\left(2zx+y, 2zy+x\right)=0$