Answer to Question #179474 in Differential Equations for kavya

Question

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

Answer 1

The given equation is incorrect, The correct equation is-

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

Taking first two terms-

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

$\Rightarrow \dfrac{dx}{x}=\dfrac{dy}{y}$

Integrating both the sides-

$logx=logy+logc_1\\ x=yc_1\\c_1=\dfrac{x}{y}~~~~~-(1)$

Taking Last two terms we have-

$\dfrac{dy}{2y(y+z^2)}=\dfrac{dz}{z^3}$

$\Rightarrow z^3dy-2y(y+z^2)dz=0$

Divide both sides by $-z^3y^2$

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

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

The above equation becomes-

$\dfrac{dy}{dz}+\dfrac{2t}{z}=\dfrac{-2}{z^3}$

On comparing it with Linear differential equation, we get-

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

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

Its solution is-

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

$\Rightarrow t(z^2)=\int z^2\times \dfrac{-2}{z^3}dz$

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

$\Rightarrow c_2=\dfrac{z^2}{y}+2logz$

The solution of the given equation is-

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