= Solution
{c}
Here $mh^2=b$, so the <Binet equation> reduces to
$$
u''+u=-\frac{(b/2)u^2e^{-u}-bu^3}{bu^2}=u-\frac12e^{-u},\qquad u''=-\frac12e^{-u}.
$$
The <initial conditions> become
$$
u(0)=\frac1a,\qquad u'(0)=-\frac1{a^2}\frac{dr}{d\theta}(0)=e^{-1/(2a)}>0.
$$
Multiply the <differential equation> by $2u'$ and use the <chain rule>:
$$
\frac{d}{d\theta}(u'^2)=-e^{-u}u'=\frac{d}{d\theta}(e^{-u}),\qquad u'^2-e^{-u}=C.
$$
The <initial conditions> give $C=0$. Since $e^{-u}>0$, $u'$ cannot change sign at a finite value of $u$. Thus $u'=e^{-u/2}$ on the physical branch. <Separation of variables> gives
$$
\frac{d}{d\theta}(e^{u/2})=\frac12,\qquad e^{u/2}=e^{1/(2a)}+\frac\theta2.
$$
Therefore
$$
\boxed{r(\theta)=\frac1{2\log\left(e^{1/(2a)}+\theta/2\right)},\qquad \theta>2\left(1-e^{1/(2a)}\right).}
$$
The <natural logarithm> must be positive because $r>0$. For the forward <trajectory>, $\theta\geq0$ and $h=\sqrt{b/m}>0$, so $\dot\theta=h/r^2>0$: the particle travels counterclockwise while $r$ strictly decreases.
For $a=1$, it starts at $(x,y)=(1,0)$, with <tangent vector> proportional to $(-e^{-1/2},1)$. Its successive crossings of the positive $x$ <coordinate axis> have radii
$$
r(2\pi k)=\frac1{2\log(\sqrt e+\pi k)},\qquad k=0,1,2,\ldots.
$$
It makes infinitely many inward turns, approaching the origin as $\theta\to\infty$. There is no finite-time arrival at the origin: $dt/d\theta=r^2/h$ behaves as $1/(4h\log^2\theta)$, whose <improper integral> diverges. If extended backwards, it comes from infinity as $\theta\downarrow2(1-\sqrt e)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-resit-orbit.png]
{title=Forward central-force trajectory for a=1, with arrows showing counterclockwise inward motion}
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