Answer to Question #179473 in Differential Equations for kavya

Question

find the general solution of the equation 2x(y+z^2)p+y(2y+z^2)q=z^3. and deduce that

yz(z^2+yz-2y)=x^2 is a solution

Answer 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

$$z=-2$$