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?
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?
Let $X=\dfrac{dR}{dt}$ ,
Then $R^3\dfrac{d}{dt}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]+4pR^2X=\dfrac{F_E}{4\pi}$
$R^3\dfrac{d}{dR}\left[\dfrac{4}{3}\rho_{\rm ext}X^2\right]\dfrac{dR}{dt}+4pR^2X=\dfrac{F_E}{4\pi}$
$\dfrac{8\rho_{\rm ext}R^3X^2}{3}\dfrac{dX}{dR}=\dfrac{F_E}{4\pi}-4pR^2X$
Let $Y=\dfrac{1}{R^2}$ ,
Then $\dfrac{dX}{dR}=\dfrac{dX}{dY}\dfrac{dY}{dR}=-\dfrac{2}{R^3}\dfrac{dX}{dY}$
$\therefore-\dfrac{16\rho_{\rm ext}X^2}{3}\dfrac{dX}{dY}=\dfrac{F_E}{4\pi}-\dfrac{4pX}{Y}$
$-\dfrac{64\pi\rho_{\rm ext}X^2}{3}\dfrac{dX}{dY}=\dfrac{F_EY-16\pi pX}{Y}$
$(F_EY-16\pi pX)\dfrac{dY}{dX}=-\dfrac{64\pi\rho_{\rm ext}X^2Y}{3}$
This belongs to an Abel equation of the second kind.
Let $U=Y-\dfrac{16\pi pX}{F_E}$ ,
Then $Y=U+\dfrac{16\pi pX}{F_E}$
$\dfrac{dY}{dX}=\dfrac{dU}{dX}+\dfrac{16\pi p}{F_E}$
$\therefore F_EU\left(\dfrac{dU}{dX}+\dfrac{16\pi p}{F_E}\right)=-\dfrac{64\pi\rho_{\rm ext}X^2}{3}\left(U+\dfrac{16\pi pX}{F_E}\right)$
$F_EU\dfrac{dU}{dX}+16\pi pU=-\dfrac{64\pi\rho_{\rm ext}X^2U}{3}-\dfrac{1024\pi^2p\rho_{\rm ext}X^3}{3F_E}$
$F_EU\dfrac{dU}{dX}=-\left(\dfrac{64\pi\rho_{\rm ext}X^2}{3}+16\pi p\right)U-\dfrac{1024\pi^2p\rho_{\rm ext}X^3}{3F_E}$