Find the angle $\Phi$ for an orbit close to a circle of radius $r$ There is an odd identity I cannot figure out in the titular exercise from V. I. Arnold's book on mechanics. He claims in an answer to the exercise (page 37) that:
\begin{align} 
\frac{M}{r^2\sqrt{V''(r)}} = \sqrt{\frac{U'}{3U'+rU''}}
\end{align}
when $V(r)=U(r)+\frac{M^2}{2r^2}$. This possibly involves a change of variables $x=M/r.$ 
I've tried to prove this but have no leads. I figure it might have something to do with the form he claims U has, although part of this chain of exercises is to find what $U$ can be. My numerical tests of the claim fail too so I must be missing something.
 A: First one has to remark that the question, while connected to the differential equation of the motion under a central force field, is not itself a differential equation but the relation between Taylor coefficients of the expansion of two potential functions.
Central force field
$\newcommand{\fx}{{\mathfrak{x}}}\newcommand{\pd}{\partial}\newcommand{\D}{{\mathit\Delta}}$
The section that this exercise is from is about the motion in the plane under a central force field
$$
\ddot\fx=-\frac{\pd U(|\fx|)}{\pd \fx}.\tag1
$$
Due to the rotational symmetry of this situation, the angular momentum is a conserved quantity, first integral, constant of motion.
After fixing the angular momentum (here $M$ instead of the more usual $L$)
$$M=r^2\dot \phi,\tag2$$
where $(r,\phi)$ are the polar coordinates of the plane point $\fx=(x,y)$, the dynamic for the radius is independent of the angle, it is
$$
\ddot r=-V'(r)\text{ where }V(r)=\frac{M^2}{2r^2}+U(r).\tag3
$$
With that as back-story, compute the derivatives of $V$
\begin{align}
V'(r) &= -\frac{M^2}{r^3}+U'(r)\tag{4a}
\\
V''(r) &= \frac{3M^2}{r^4}+U''(r)\tag{4b}
\end{align}
Perturbation of a circular orbit

The exercise in question now asks to find the angle $\Phi$ between pericenter and apocenter for an orbit close to a circular orbit of radius $r^*$ (and the same angular momentum).

Previously it was explained that the radius of such a circular orbit is at a (regular) minimum of $V$, so that $V'(r^*)=0$, $V''(r^*)>0$. The perturbed orbit has an energy $$\frac12\dot r^2+V(r)=E=V(r^*)+\D E$$ so that at the extremal points with $\dot r=0$ one gets in the Taylor expansions around $r^*$ the equation $$V(r^*)+\D E=V(r^*+ \D r)=V(r^*)+\tfrac12 V''(r^*)(\D r)^2+O((\D r)^3).$$ Simplifying and solving for the increment gives $$r_{\max},r_{\min}\approx r^*\pm\D r$$ with $$\D r=\sqrt{\frac{2\D E}{V''(r^*)}}. $$
Half-period
By the previous exercise, the half-period is computed as
\begin{align}
\Phi&=\int_{r_{\min}}^{r_{\max}}\frac{M/r^2}{\sqrt{2(E-V(r))}}dr\tag{5a}
\\
&\approx\int_{r_{\min}}^{r_{\max}}\frac{M/r^2}{\sqrt{2(\D E-\frac12V''(r^*)(r-r^*)^2)}}dr
\\
&=\int_{-1}^{1}\frac{M/(r^*+s\D r)^2}{\sqrt{V''(r^*)}\sqrt{1-s^2}}ds
=\frac{M/(r^*)^2}{\sqrt{V''(r^*)}} \int_{-1}^{1}\frac1{\sqrt{1-s^2}}ds+O((\D r)^2)
\\
&\approx\Phi_{\rm cir}=\pi\frac{M/(r^*)^2}{\sqrt{V''(r^*)}}\tag{5b}
\end{align}
Now use $V'(r^*)=0\implies U'(r^*)=\frac{M^2}{(r^*)^3}$ to eliminate the auxiliary variables $M$ and $V$ from this formula (not completely, as $r^*$ depends on $M$) to get
$$
V''(r^*)=\frac{3}{r^*}U'(r^*)+U''(r^*)
\\
\frac{M}{(r^*)^2}=\sqrt{\frac{U'(r^*)}{r^*}}
\\
\implies
\Phi_{\rm cir}=\pi\frac{M/(r^*)^2}{\sqrt{V''(r^*)}}=\pi\sqrt{\frac{U'(r^*)}{3U'(r^*)+r^*U''(r^*)}}.\tag6
$$
A: $$\frac{M}{r^2\sqrt{V''(r)}} = \sqrt{\frac{U'}{3U'+rU''}}$$
$$\frac{M^2}{r^4V''(r)} = \frac{U'}{3U'+rU''}$$
$$\frac{r^4}{M^2}V''(r)=3+r\frac{U''}{U'}$$
Squaring might introduce additional solutions. At end of this calculus one must check and keep only the valid solutions.
$$V(r)=U(r)+\frac{M^2}{2r^2}$$
$$V''(r)=U''(r)+\frac{3M^2}{r^4}$$
$$\frac{r^4}{M^2}(U''(r)+\frac{3M^2}{r^4})=3+r\frac{U''}{U'}$$
$$(\frac{r^4}{M^2}-r\frac{1}{U'})U''(r)=0$$
First set of solutions , from $U''(r)=0$ :
$$\boxed{\begin{cases}
U(r)=c_1r+c_2\\ 
V(r)=c_1r+c_2+\frac{M^2}{2r^2}
\end{cases}}$$
Second set of solutions, from $\frac{r^4}{M^2}-r\frac{1}{U'}=0\quad\implies\quad U'=\frac{M^2}{r^3}$
$$\begin{cases}
U(r)=-\frac{M^2}{2r^2}+c_3\\ 
V(r)=-\frac{M^2}{2r^2}+c_3+\frac{M^2}{2r^2}=c_3
\end{cases}\quad\text{rejected because}\quad \begin{cases}
V''(r)=0\\
3U'+rU''=0
\end{cases}$$
The original ODE becomes $\quad\frac{M}{r^2\sqrt{0}} = \sqrt{\frac{U'}{0}}=\infty$
This is the conclusion in a mathematical sense. Nevertheless one should check whether the solution $\quad \begin{cases}
U(r)=-\frac{M^2}{2r^2}+c_3\\ 
V(r)=c_3
\end{cases}\quad$ has a physical meaning.
