Answer to Question #147413 in Differential Equations for Vedant

Question

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

Answer 1

P=y-zx, Q=x+yz, R=x²+y²

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

Multiplyers: -y,-x,1

$\frac{-ydx-xdy+dz}{-y^2+xyz-x^2-xyz+x^2+y^2}=0$

-ydx-xdy+dz=0

-yx-xy+z=c₁

c₁=z-2xy

Multiplyers: y,x,-1

$\frac{ydx+xdy-dz}{y^2-zxy+x^2+zxy-x^2-y^2}=0$

ydx+xdy-dz=0

yx+xy-z=c₂

c₂=2xy-z

Answer: c₁=z-2xy, c₂=2xy-z