= Solution
The <Euler-Lagrange equations> of the <particle Lagrangian in a shearing sheet> are
$$
\ddot x-2\Omega_0\dot y-2q_0\Omega_0^2x=0,\qquad
\ddot y+2\Omega_0\dot x=0,\qquad
\ddot z+\Omega_z^2z=0.
$$
The terms coupling $x$ and $y$ are the <Coriolis acceleration>. The <cyclic coordinate> $y$ has conserved <canonical momentum>
$$
p_y=\frac{\partial L_2}{\partial\dot y}=\dot y+2\Omega_0x.
$$
For a <Newtonian potential of a point mass>, $q_0=3/2$ and $\Omega_z=\kappa_r=\Omega_0$. Set $p_y=\Omega_0x_0/2$ and substitute $\dot y=p_y-2\Omega_0x$ in the radial equation. It becomes $\ddot x+\Omega_0^2(x-x_0)=0$, a <harmonic oscillator equation> about the <epicyclic guiding center> $x_0$. Integration gives
$$
\boxed{\begin{aligned}
x&=x_0+\operatorname{Re}(Ae^{-i\Omega_0t}),\\
y&=y_0-\frac32\Omega_0x_0t+\operatorname{Re}(-2iAe^{-i\Omega_0t}),\\
z&=\operatorname{Re}(Be^{-i\Omega_0t}).
\end{aligned}}
$$
The four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.
Expanding the inertial <specific angular momentum> gives $h=h_0+r_0p_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference <circular orbit>, and $x_0$ specifies the radius of its associated <epicyclic guiding center>.
The conserved horizontal energy in the rotating frame is
$$
\varepsilon_h=\frac12(\dot x^2+\dot y^2)-\frac32\Omega_0^2x^2
=\boxed{\frac12\Omega_0^2\left(|A|^2-\frac34x_0^2\right)}.
$$
It is the horizontal part of the local <Jacobi energy in a shearing sheet>, rather than the inertial <specific orbital energy>. Its positive term is the <epicyclic energy>; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical <harmonic oscillator> has conserved energy
$$
\varepsilon_v=\frac12(\dot z^2+\Omega_0^2z^2)
=\boxed{\frac12\Omega_0^2|B|^2}.
$$
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