Question

Find general solution of 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

$$Solution$$

$$2x(y+z^2)\frac{dz}{dx}+y(2y+z^2)\frac{dz}{dy}=z^2$$

$$\frac{dx}{2x(y+z^2)}=\frac{dy}{y(2y+z^2)}=\frac{dz}{z^2}$$

$$y\frac{dx}{2xy^2+2xyz^2}=x\frac{dy}{2xy^2+xyz^2}=xy\frac{dz}{xyz^2}$$

$$\frac{ydx−xdy−xydz}{2xy2+2xyz2−2xy2−xyz2−xyz2}=\frac{ydx−xdy−xydz}{0}$$

$$ydx−xdy−xydz=0$$

$$\frac{dx}{x}−\frac{dy}{y}−dz=0$$

$$lnx−lny−z=c$$

$$z=ln( \frac{x}{y})+C$$

2.)

$$yz(z^2+2z−2y)=x^2$$

$$yz^3+2yz^2−2y2z=x^2$$

Differentiate respect to x:

$$3yz^2p+4yzp−2y^2p=2x$$

$$py(3z^2+4z−2y)=2x$$

Differentiate respect to y:

$$z^3+3z^2yq+2z^2+4yzq−4yz−2y^2q=0$$

$$qy(3z^2+4z−2y)+z^3+2z^2−4yz=0$$

$\therefore$ we have:

$$3z^2+4z−2y=\frac{2x}{py}$$

$$\frac{2xq}{p}+z^3+2z^2−4yz=0$$

Since

$$p=\frac{1}{x}$$

$$q=\frac{−1}{y}$$

then:

$$\frac{−2x^2}{y}+z^3+2z^2−4yz=0$$

$$−2x^2+y(z^3+2z^2−4yz)=0$$

$$yz(z^2+2z−4y)=2x^2$$

So:

$$\frac{z^2+2z−4y}{z^2+2z−2y}=2$$

$$z^2+2z−4y=2z^2+4z−4y$$

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

So we get that the statement can be proved if

$$z=0, or$$

$$z=−2$$