Question

Z^2dx+(z^2-2yz)dy+(2y^2-yz-xz) dz=0

Answer 1

Comparing the given equation with the standard equation $Pdx+Qdy+Rdz=0,$ we get

$P=z^2,Q=z^2-2yz,R=2y^2-yz-xz$

The given equation is homogeneous of degree 2. Now first of all, we test the condition of integrability.

$P\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)-Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ =z^2(4y-z-2z+2y)-(z^2-2yz)(-z-2z)+(2y^2-yz-xz)(0-0)\\ =6yz^2-3z^3+3z^3-6yz^2=0$

Hence, the equation is integrable.

Substitute $x=uz,y=vz$ into the equation.

This implies that, $dx=udz+zdu,dy=vdz+zdv$

$z^2(udz+zdu)+z^2(1-2v)(vdz+zdv)+z^2(2v^2-v-u)dz=0$

With this, the coefficient of $dz$ will be 0. SO, we have;

$z^3du+z^3(1-2v)dv=0$

Divide through by $z^3$

$du+(1-2v)dv=0$

Integrate through

$u+v-v^2=C$

But, $u=\frac{x}{z},y=\frac{y}{z}$

$\frac{x}{z}+\frac{y}{x}-\frac{y^2}{z^2}=C\\ xz+yz-y^2=Cz^2$

Hence, the general solution of the equation is $xz+yz-y^2=Cz^2$ .