# Analytical solution of $\ddot r=ar^{-2}+b(\dot{r})^2$

I have this physics problem I'm trying to solve and its been a while since I've done differential equautions. The problem I'm trying to solve is:

$$-GMm/r^2 -k(\dot r)^2 = m\ddot r.$$

I know that I can assume that $r = e^{pt}$, sub that in and then possibly taylor expand $e^{pt}$ to approximate my $p$ values, but the solution I see in the book looks like it could be analytical.

I'm playing around with equations that look like $\ln(r)$ to see if I can get an analytical solution that fits the bill, but to no avail.

So my question, is there a classification of ode's of the following form with analytical solutions I can look up?

$$\ddot r=ar^{-2}+b\dot{r}^2$$

• For what it's worth, Wolfram Alpha can't find an analytical solution to this form (except if $k=0$, but even then the solution is very complex). Jul 1 '16 at 10:13
• I don't understand the writing, you mean $ar^{-2}+b\dot{r}^2=c\ddot{r}$ ? Jul 1 '16 at 11:35
• Indeed the notations $dr$ and $d^2r$ need serious explanations.
– Did
Jul 1 '16 at 12:25
• @user90369 thats correct. Sorry I wrote this on my phone. Jul 1 '16 at 12:29
• @Dr.Knowitall: Then search for autonomous differential equation: $y''=f(y,y')$. Jul 1 '16 at 12:32

$$\frac{d^2r}{dt^2}=\frac{a}{r^2}+b\left(\frac{dr}{dt} \right)^2$$ This is an ODE of the autonomous kind. The usual method to solve if is the change of function :
$$y(r)=\frac{dr}{dt} \quad\to\quad \frac{d^2r}{dt^2}=\frac{dy}{dt}=\frac{dy}{dr}\frac{dr}{dt}=y\frac{dy}{dr}$$ This transformes the ODE into : $$y\frac{dy}{dr}=\frac{a}{r^2}+by^2$$ Let $Y(r)=y^2$ $$\frac{1}{2}\frac{dY}{dr}=\frac{a}{r^2}+bY \quad\to\quad \frac{dY}{dr}-2bY=\frac{2a}{r^2}$$ This is a first order linear nonhomogeneous ODE.
Solving the associated homogeneous ODE : $\quad \frac{dY}{dr}-2bY=0\quad$ leads to $\quad Y=c\:e^{2br}\quad$ so we search the solution of the inhomogeneous ODE on the form $\quad Y=f(r)e^{2br}$
$\frac{dY}{dr}-2bY=\frac{2a}{r^2} = \frac{df}{dr}e^{2br}+2bf(r)e^{2br}-2bf(r)e^{2br} =\frac{df}{dr}e^{2br}$ $$\frac{df}{dr}=\frac{2a}{r^2}e^{-2br}$$ $$f(r)=2a\int \frac{e^{-2br}}{r^2}dr = 2a\left(-\frac{e^{-2br}}{r} -2b\text{Ei}(-2br)\right)+\text{constant}$$ Ei is the special function named Exponential Integral : http://mathworld.wolfram.com/ExponentialIntegral.html $$Y(r)=f(r)e^{2br}= -\frac{2a}{r} -4ab\:\text{Ei}(-2br)+c_1e^{2br}$$ $$y(r)=\pm\sqrt{-\frac{2a}{r} -4ab\:\text{Ei}(-2br)+c_1e^{2br} }$$ $y(r)=\frac{dr}{dt} \quad\to\quad t=\int \frac{dr}{y(r)}$ $$t=\pm\int \frac{dr}{\sqrt{-\frac{2a}{r} -4ab\:\text{Ei}(-2br)+c_1e^{2br} } }+c_2$$ This is the general solution of the ODE, expressed on implicit form $t$ as a function of $r$.
There is no closed form for the integral. So, a-fortiori, no closed form for the inverse function $r(t)$.