Answer to Question #171854 in Differential Equations for J. Sandilya.

Question

(y+z)dx+(z+x)dy+(x+y)dz=0 find the solution

Answer 1

Solve:

$(y+z)dx+(z+x)dy+(x+y)dz=0.$

Solution:

Here, $P=y+z,Q=z+x,R=x+y.$

$\because \sum P(\frac{\partial{Q} }{\partial{z}}-\frac{\partial{R}}{\partial{y}})=0.$

condition of integrability satisfed.

Rearranging the terms of the given equation,

$(ydx+xdy)+(zdx+xdz)+(zdy+ydz)=0$

$\Rarr d(xy)+d(zx)+d(zy)=0.$

on integration

Answer:

$xy+yz+zx=C.$