Question #112515
(Yz+2x)dx+(zx+2y)dy+(xy+2z)dz=0
1
Expert's answer
2020-04-29T16:17:46-0400

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

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

2xdx+2ydy+2zdz=d(x2+y2+z2)2xdx+2ydy+2zdz=d(x^2+y^2+z^2)

Then d(x2+y2+z2+xyz)=0d(x^2+y^2+z^2+xyz)=0

So x2+y2+z2+xyz=Cx^2+y^2+z^2+xyz=C

Answer: x2+y2+z2+xyz=Cx^2+y^2+z^2+xyz=C


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