Answer to Question #181336 in Differential Equations for kamal

Question

solve dx/2x(y+z^2) = dy/y(2y+z^2) = dz/z^3

Answer 1

The correct form of the given equation is:

$$\frac{dx}{2x(y+z^2)}=\frac{dy}{2y(y+z^2)}=\frac{dz}{z^3}$$

Taking the first two terms

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

Integrating both the sides, we have:

$$\log x=\log y + \log c_1\\ x=yc_1\\ c_1 = \frac{x}{y} \qquad \cdots (1)$$

Taking Last two terms, we have:

$$\frac{dx}{2y(y+z^2)}=\frac{dz}{z^3}\\ \implies z^3dy-2y(y+z^2)dz = 0$$

Divide both sides by $-z^3y^2$

$$-\frac{1}{y^2}\frac{dy}{dz}+\frac{2}{zy}=-\frac{2}{z^3}\\$$

Let

$$\frac{1}{y} =t \\ \implies -\frac{1}{y^2}\frac{dy}{dz} =\frac{dt}{dz}$$

On comparing it with Linear differential equation, we get:

$$P= \frac{2}{z} ,Q= \frac{-2}{z^3}$$

Integrating factor, I.F.=$e^{\int P.dz}=e^{\int \tfrac{2}{z}dz}=e^{2logz}=z^2$

The solution is:

$$t\times I.F.=\int I.F.\times Q dz$$

$$⇒t(z^2 )=∫z^2×\frac{-2}{z^3}dz$$

$$\Rightarrow t\times z^2=-2logz+c_2\\ ​$$

$$⇒c_2= \frac{z^2}{y}+2logz$$

The solution of the given equation is:

$$\phi(c_1,c_2)=0 \implies \phi(\dfrac{x}{y},\dfrac{z^2}{y}+2logz)=0$$