With a potential which is proportional to the inverse square of the radius, you can write the energy as
$$E=\frac{1}{2}m \dot{r}^2 +\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
where $L$ is the angular momentum
$$L=m r^2 \dot{\phi}$$
You can rewrite the energy in the following way
$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \dot{\phi} \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
Now substitute $\dot{\phi}$ using the definition of the angular momentum
$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \frac{L}{mr^2} \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
and change variable setting $u=1/r$. You obtain
$$E=\frac{L^2}{2m}\left( \frac{du}{d\phi} \right)^2+\left( \frac{L^2}{2m}-a\right)u^2$$
By setting $dE/d\phi=0$ (energy is conserved) you obtain a motion equation of the form
$$\frac{d^2u}{d\phi^2}+\left(1-\frac{2ma}{L^2} \right)u=0$$
If $2ma/L^2<1$ this is an harmonic oscillator, and the general solution is
$$u=\frac{1}{r} = A\cos \left( k \phi + B \right)$$ with
$$k=\sqrt{1-\frac{2ma}{L^2}}$$ and $A$,$B$ arbitrary constants.
This describes an orbit that winds $1/k$ times around the attraction centre and then escape to the infinity. If $2ma/L^2>1$ the potential is larger that the centrifugal barrier. The solution of the equation can be written as
$$u=\frac{1}{r} = A\cosh \left( k \phi + B \right)$$ and the particle falls in the centre of attraction (with an infinite number of windings).
The particular case $2ma/L^2=1$ correspond to
$$u=\frac{1}{r} = A \phi + B$$
In this case the potential cancels exactly the centrifugal barrier, and the particle falls on the centre of attraction coming from the infinity.
There is an interesting geometrical interpretation of these results when $k$ is real. If you introduce
$$\tilde{L} = L \sqrt{1-\frac{2ma}{L^2}}$$
you can rewrite the energy as
$$E=\frac{1}{2}m \dot{r}^2 + \frac{\tilde{L}^2}{2mr^2}$$
and
$$\tilde{L} = mr^2 \dot{\theta}$$
where the new "angular variable" is
$$\theta= \sqrt{1-\frac{2ma}{L^2}} \phi$$
$E$ and $\tilde{L}$ seems to be the energy and the angular momentum of a free particle, in polar coordinates. The trajectories will be simply lines.
However a complete orbit around the centre $\Delta \phi=2\pi$ is given by
$$\Delta \theta = 2\pi \sqrt{1-\frac{2ma}{L^2}}$$
so you must cut the plane at $\theta= \Delta \theta$ and identify the border with $\theta=0$ (in other words, you obtain a cone). The "line" will wind around the attraction centre several times, if the factor $k=\sqrt{1-\frac{2ma}{L^2}}$ is small, accordingly with the analytic solution.
