Answer to Question #86758 in Differential Equations for Kalipada

Question

2x(y+z^2)p+y(2y+z^2)q=z^2
And prove thatyz(z^2+2z-2y)=x^2

Answer 1

1)

$$2x(y+z^2)\partial z/\partial x+y(2y+z^2)\partial z/\partial y=z^2$$

$$dx/(2x(y+z^2))=dy/(y(2y+z^2))=dz/z^2$$

$$ydx/(2xy^2+2xyz^2)=xdy/(2xy^2+xyz^2))=xydz/(xyz^2)$$

$$(ydx-xdy-xydz)/(2xy^2+2xyz^2-2xy^2-xyz^2-xyz^2)=(ydx-xdy-xydz)/0$$

$$ydx-xdy-xydz=0$$

$$dx/x-dy/y-dz=0$$

$$lnx-lny-z=c$$

$$z=ln(x/y)+C$$

2)

$$yz(z^2+2z-2y)=x^2$$

$$yz^3+2yz^2-2y^2z=x^2$$

Differentiate respect to x:

$$3yz^2p+4yzp-2y^2p=2x$$

$$py(3z^2+4z-2y)=2x$$

Differentiate respect to y:

$$z^3+3z^2yq+2z^2+4yzq-4yz-2y^2q=0$$

$$qy(3z^2+4z-2y)+z^3+2z^2-4yz=0$$

So we have:

$$3z^2+4z-2y=2x/(py)$$

$$2xq/p+z^3+2z^2-4yz=0$$

Since

$$p=1/x$$

$$q=-1/y$$

then:

$$-2x^2/y+z^3+2z^2-4yz=0$$

$$-2x^2+y(z^3+2z^2-4yz)=0$$

$$yz(z^2+2z-4y)=2x^2$$

So:

$$(z^2+2z-4y)/(z^2+2z-2y)=2$$

$$z^2+2z-4y=2z^2+4z-4y$$

$$z^2+2z=0$$

So we get that the statement can be proved if

$$z=0$$

or

$$z=-2$$