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I need to solve the equation \begin{eqnarray} R^3 \frac{d } {dt} \left [ \frac{4}{3} \rho_{\rm ext} \left ( \frac{dR}{dt} \right )^2 \right ]+ 4 p R^2 \frac{d R} {dt} =\frac{F_E}{4\pi} \end{eqnarray}

Could you please help in this regard?

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migrated from Sep 6 '12 at 16:20

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Let $X=\dfrac{dR}{dt}$ ,

Then $R^3\dfrac{d}{dt}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]+4pR^2X=\dfrac{F_E}{4\pi}$

$R^3\dfrac{d}{dR}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]\dfrac{dR}{dt}+4pR^2X=\dfrac{F_E}{4\pi}$

$\dfrac{8\rho_{\rm ext}R^3X^2}{3}\dfrac{dX}{dR}=\dfrac{F_E}{4\pi}-4pR^2X$

$X^2\dfrac{dX}{dR}=\dfrac{3F_E}{32\pi\rho_{\rm ext}R^3}-\dfrac{3pX}{2\rho_{\rm ext}R}$

Let $X=\dfrac{1}{Y}$ ,

Then $\dfrac{dX}{dR}=-\dfrac{1}{Y^2}\dfrac{dY}{dR}$

$\therefore-\dfrac{1}{Y^4}\dfrac{dY}{dR}=\dfrac{3F_E}{32\pi\rho_{\rm ext}R^3}-\dfrac{3p}{2\rho_{\rm ext}RY}$

$\dfrac{dY}{dR}=-\dfrac{3F_EY^4}{32\pi\rho_{\rm ext}R^3}+\dfrac{3pY^3}{2\rho_{\rm ext}R}$

This belongs to a "Chini-like" equation as mentioned here and which is more complicated than Abel equation of the first kind.

I still don't know whether existing a substitution so that $Y^3$ term can be eliminated and keeping $Y^2$ term still vanished, leading it exactly belongs to a Chini equation.

It still don't know whether existing method can solve Chini equation generally, while Abel equation of the first kind eventually can be solved generally starting in 2011 August, see this for details.

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