I suddenly became curious about the following differential equation: \begin{align*} \frac{d^2x}{dt^2} = \frac{k}{x(t)^2} && x(0) = x_0 > 0 && \frac{dx}{dt}(0) = v_0 \end{align*} the "1D inverse square law problem".
It's a pretty natural thing to think about. You're a positive distance $x_0$ away from a point particle which is exerting an inverse square force on you. You're moving only radially with respect to that particle with initial velocity $v_0$. What happens? Do all solutions have a "closed form"? Do some of them?
Obviously the behaviour can be quite different depending on the intial conditions. For example, if $k <0$, you're either going to fall into the "black hole", or you're going to escape to infinity (or is it even possible to asympotically approach some finite maximum distance from below as $t \to +\infty$?).
Edit: I thought of one solution: letting $x(t) = a t^{2/3}$, one gets $\frac{d^2x}{dt^2} = \frac{-2a}{9} t^{-4/3}$ so that $x^2 \cdot \frac{d^2 x}{dt^2} = \frac{-2a^3}{9}$ whence, taking $a = - \left( \frac{9k}{2} \right)^{1/3}$, we get a solution. We could also shift the time variable to get some more solutions.
Edit2: The preceding solution is simple enough that I figured this was the "minimal escape curve" i.e. the one where you have precisely enough energy to escape to infinity. Indeed, it's not hard to check this by calculuating the kinetic and potential energy when $t= 1$ i.e. $x=a$. But my friend pointed out a much nicer way to see this! Notice $\frac{dx}{dt} = \frac{2a}{3} t^{-1/3} \to 0$ as $t \to \infty$. Since the veclocity goes to zero, the kinetic energy goes to zero. So, by conservation of energy, this is the minimal escape curve (if you had any spare energy left at $x = \infty$, you wouldn't have slowed to velocity zero).
Edit3: In the end I found out how to get the solutions implicitly. See my answer here.
See this answer at Physics SE.
In short, this attractive inverse squares problem is somewhat ill-defined in 1 dimension.
If you look at your problem as limit case of Kepler problem with angular momentum $M\to0$, then you have the particle moving from its initial position to the singularity of the potential, and then bouncing off it (because the ellipse with one focus at singularity will degenerate into a line with one of its ends at singularity).
If, on the other hand, you look at the problem as initially 1-dimensional and the potential being the limit of $U=\frac{k}{\sqrt{x^2+\varepsilon^2}}$ as $\varepsilon\to0$, then your particle would go through the singularity, contrary to bouncing off it.
As these limits, having identical final potential, give different results, the original problem doesn't have a consistent solution on its own.