= Solution
Write $z=\sqrt\epsilon\,\omega\cos\psi$, with $\omega=\sqrt\beta>0$ and $\psi=\phi-\phi_0$. Differentiating twice gives
$$
\frac{z''+kz}{\sqrt\epsilon}
=[\omega''-\omega(\phi')^2+k\omega]\cos\psi
-[2\omega'\phi'+\omega\phi'']\sin\psi.
$$
The coefficients must vanish for arbitrary initial phase $\phi_0$. The sine coefficient gives
$$
(\omega^2\phi')'=0,\qquad \omega^2\phi'=C.
$$
The conventional <Courant-Snyder parameters> normalize the phase so that $C=1$; any positive constant $C$ can be absorbed into the scale assigned to $\beta$ and $\epsilon$. Explicitly, $\widetilde\beta=\beta/C$ and $\widetilde\epsilon=C\epsilon$ preserve the orbit while making $\widetilde\beta\phi'=1$. With this convention,
$$
\boxed{\phi'=\frac1\beta,\qquad
\omega''-\frac1{\omega^3}+k(s)\omega=0.}
$$
The second expression is the <Ermakov-Pinney equation> for the beam envelope. It follows by substituting $\phi'=1/\omega^2$ into the cosine coefficient.
\b[The printed equation omits the factor $\omega$ multiplying $k(s)$.] For a concrete counterexample, take the constant periodic focusing strength $k=1/16$ and the constant envelope $\omega=2$. Then $\phi'=1/4$ and $z=2\sqrt\epsilon\cos(s/4-\phi_0)$ solve $z''+z/16=0$. The correct envelope equation is $0-1/8+(1/16)2=0$, whereas the printed equation gives $-1/8+1/16=-1/16$.
The associated invariant is
$$
\epsilon=\gamma_Tz^2+2\alpha_Tzz'+\beta z'^2,
\qquad \alpha_T=-\beta'/2,\quad
\gamma_T=(1+\alpha_T^2)/\beta.
$$
Thus $\epsilon$ fixes the <phase space>[phase-space] ellipse of a trajectory; a beam's rms <geometric emittance> uses the corresponding second-moment convention. Factors of $\pi$ depend on whether emittance means ellipse area or area divided by $\pi$ and do not alter the envelope equation.
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