Closed form solution for $\rho$: $\frac{d}{d\tau}\sqrt{\left(\rho^{3}\frac{d^{2}\rho}{d\tau^{2}}+\rho\right)}=\nu\rho$ with initial conditions

Question

Consider the following differential equation for $\rho$: $$\frac{d}{d\tau}\sqrt{\left(\rho^{3}\frac{d^{2}\rho}{d\tau^{2}}+\rho\right)}=\nu\rho$$ This equation can be rewritten as a system of 3 first-order differential equations: $$\frac{d\rho}{d\tau}=A,$$ $$\frac{dA}{d\tau}=\frac{B^{2}-\rho}{\rho^{3}},$$ $$\frac{dB}{d\tau}=\nu\rho.$$ Therefore, it is possible to numerically solve this equation eg using a Runge-Kutta method. However, I would like to know if it is possible to find a closed form solution to this equation with initial conditions: $\rho=1$, $A=0$ and $B=1$. Any help is welcome.

Answer 1

Hint:

Let $u=\dfrac{d\rho}{d\tau}$ ,

Then $\dfrac{d^2\rho}{d\tau^2}=\dfrac{du}{d\tau}=\dfrac{du}{d\rho}\dfrac{d\rho}{d\tau}=u\dfrac{du}{d\rho}$

$\therefore\dfrac{d}{d\rho}\left(\sqrt{\rho^3u\dfrac{du}{d\rho}+\rho}\right)\dfrac{d\rho}{d\tau}=\nu\rho$

$\dfrac{u\left(\rho^3u\dfrac{d^2u}{d\rho^2}+\rho^3\left(\dfrac{du}{d\rho}\right)^2+3\rho^2u\dfrac{du}{d\rho}+1\right)}{2\sqrt{\rho^3u\dfrac{du}{d\rho}+\rho}}=\nu\rho$