Answer to Question #179785 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

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

$$y\frac{dx}{2xy^2+2xyz^2}=x\frac{dy}{2xy^2+xyz^2}=xy\frac{dz}{xyz^2}$$ $$\frac{ydx−xdy−xydz}{2xy2+2xyz2−2xy2−xyz2−xyz2}=\frac{ydx−xdy−xydz}{0}$$ $$ydx−xdy−xydz=0$$ $$\frac{dx}{x}−\frac{dy}{y}−dz=0$$ $$lnx−lny−z=c$$ $$z=ln( \frac{x}{y})+C$$

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

$$z=0, or$$ $$z=−2$$