= Solution
The relativistic orbit-shape equation can be derived without dividing by $\dot r$ at a turning point. Substitute the norm constraint into the radial <geodesic equation> to eliminate $\dot t$ and $\dot r^2$. Using $f'=2\mu/r^2$ gives
$$
\ddot r=-\frac{\mu\kappa}{r^2}+\frac{h^2}{r^3}-\frac{3\mu h^2}{r^4}.
$$
For a massive particle with <proper time> as parameter, $\kappa=c^2$. Put $u=1/r$, and use primes in this paragraph for <derivatives> with respect to $\phi$. The constant <specific angular momentum> gives $\dot\phi=hu^2$, so the <chain rule> yields
$$
\dot r=\frac{dr}{d\phi}\dot\phi=-hu',\qquad
\ddot r=-h^2u^2u''.
$$
Substituting these expressions into the radial equation and dividing by $h^2u^2$ for a nonradial orbit gives
$$
\boxed{u''+u-3\mu u^2=\frac{\mu c^2}{h^2}=\frac{GM}{h^2}.}
$$
The extra $3\mu u^2$ term is responsible for <Schwarzschild perihelion precession>. It vanishes in the Newtonian weak-field limit $\mu/r\to0$. The derivation uses a massive particle; for a <null geodesic>, $\kappa=0$ and the constant $GM/h^2$ on the right is absent.
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