= Solution
In <Newtonian gravity>, the radial equation with conserved <specific angular momentum> is $\ddot r=-GM/r^2+h^2/r^3$. The substitution $u=1/r$ therefore gives the linear <differential equation>
$$
u''+u=\frac{GM}{h^2}.
$$
Choosing $\phi=0$ along perihelion, its <elliptic orbit> solution is
$$
\boxed{u_0(\phi)=\frac{1+e\cos\phi}{p},\qquad
p=\frac{h^2}{GM}=a_{\rm orb}(1-e^2).}
$$
Here $p$ is the <semilatus rectum>. One can check the last equality directly from $r_{\rm peri}=p/(1+e)$, $r_{\rm apo}=p/(1-e)$, and $a_{\rm orb}=(r_{\rm peri}+r_{\rm apo})/2$.
To find <Schwarzschild perihelion precession>, write $u=u_0+\delta u$ in the relativistic equation and retain first order in $\mu/p$. The <linear ordinary differential equation> for the correction is
$$
\delta u''+\delta u=3\mu u_0^2
=\frac{3\mu}{p^2}\left(1+\frac{e^2}{2}+2e\cos\phi+\frac{e^2}{2}\cos2\phi\right).
$$
The constant and double-frequency forcing change the orbit's shape without accumulating an angular shift. The resonant $\cos\phi$ term produces
$$
\delta u_{\rm sec}=\frac{3\mu e}{p^2}\,\phi\sin\phi,
$$
since $(d^2/d\phi^2+1)(\phi\sin\phi)=2\cos\phi$. The oscillating part of the perturbed orbit is consequently the first-order expansion of
$$
\frac e p\cos\big[(1-3\mu/p)\phi\big].
$$
Successive perihelia are separated by $2\pi/(1-3\mu/p)$, slightly more than a full revolution. Their forward angular advance is thus
$$
\boxed{\Delta\phi=\frac{6\pi\mu}{p}
=\frac{6\pi GM}{c^2a_{\rm orb}(1-e^2)}
+O\!\left(\frac{\mu^2}{p^2}\right)}
$$
in radians per orbit. This leading expression applies when $\mu/p\ll1$. For a perfectly <circular orbit> there is no uniquely located perihelion; the same expression describes the limiting mismatch of radial and azimuthal frequencies for a small nonzero <orbital eccentricity>.
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