= Solution
{c}
The initial transverse <velocity> gives $h=a(L/a)=L$. From $\dot r=-hu'$, the <initial conditions> are $u(0)=1/a$ and $u'(0)=1/a$. Define
$$
\nu=\sqrt{1+L^{-2}}>1.
$$
The <Binet equation> becomes the constant-coefficient <linear ordinary differential equation>
$$
u''+\nu^2u=\frac1{aL^2}.
$$
Its <particular solution> is $(\nu^2-1)/(a\nu^2)$. Applying the two <initial conditions> to the <complementary solution> gives
$$
\boxed{\frac1r=\frac{\nu^2-1+\cos(\nu\theta)+\nu\sin(\nu\theta)}{a\nu^2}.}
$$
The physical branch is the connected range containing $\theta=0$ on which the right-hand side is positive. A zero corresponds to escape to infinite <radius>, rather than continuation to negative $r$.
The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(ar^3)$ changes sign at $r=a$. The stated <initial condition> has inward radial <velocity>, so the particle immediately enters $r<a$, where the <force> is outward. The consistent interpretation is that $F$ is the \b[signed inward radial component] of the <force> per unit <mass>.
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