= Solution
To first order in $r_0$,
$$
\frac{K}{(1-r_0u)^2}=K(1+2r_0u)+O(r_0^2).
$$
The expanded orbit equation is therefore
$$
u''+\omega^2u=K,
\qquad
\omega^2=1-2Kr_0.
$$
Its required family of solutions is
$$
\boxed{
u(\theta)=X[1+A\cos(\omega\theta)]},
\qquad
\boxed{X=\frac{K}{1-2Kr_0},\quad
\omega=\sqrt{1-2Kr_0}},
$$
with any $0<A<1$. Successive periapses occur when $\omega\theta$ changes by $2\pi$, so their angular separation is
$$
\boxed{\Delta\theta=\frac{2\pi}{\sqrt{1-2Kr_0}}
=2\pi(1+Kr_0)+O(r_0^2)}.
$$
The periapsis advance per revolution is consequently $2\pi Kr_0+O(r_0^2)$.
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