Question

(xz+yz)dx/dy + (xz-yz)dz/dy = x²+y² find the general solution

Answer 1

(xz+yz)p+(xz-yz)q=x²+y²

P=xz+yz

Q=xz-yz

R=x²+y²

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

$\frac{dx}{xz+yz}= \frac{dy}{xz-yz}=\frac{dz}{x^2+y^2}$

Multipliers: -x,y,z

-xdx+ydy+zdz=0

1/2(-x²+y²+z²)=C₁

Multipliers: x,-y,-z

xdx-ydy-zdz=0

1/2(x²-y²-z²)=C₂

$\phi(c_{1},c_{2})=0$

$\phi( 1/2(-x^2+y^2+z^2),1/2(x^2-y^2-z^2))=0$