Answer to Question #112515 in Differential Equations for Keerthi

Question

(Yz+2x)dx+(zx+2y)dy+(xy+2z)dz=0

Answer 1

We can rewrite the equation: $(yzdx+zxdy+xydz)+(2xdx+2ydy+2zdz)=0$

$yzdx+zxdy+xydz=d(xyz)$

$2xdx+2ydy+2zdz=d(x^2+y^2+z^2)$

Then $d(x^2+y^2+z^2+xyz)=0$

So $x^2+y^2+z^2+xyz=C$

Answer: $x^2+y^2+z^2+xyz=C$