= Solution
For either scalar equation, the <complementary solution> is
$$
v'_{x,h}=e^{-\epsilon\Omega t}\bigl[A(x)\cos(\Omega t)+B(x)\sin(\Omega t)\bigr],
$$
so initial velocity transients decay when $t\gg(\epsilon\Omega)^{-1}$. In the exact corrected <forced dust epicycle with aerodynamic drag>, $v'_{x,p}=u\cos\theta$ is a <particular solution>, and the original radial equation then gives $v'_{y,p}=u\sin\theta/2$. Consequently
$$
\boxed{\mathbf v'\longrightarrow\mathbf u'\quad\text{at late times}.}
$$
This is <resonant dust entrainment by an epicyclic gas wave>: in the adopted long-wavelength gas approximation, the forcing is a free epicycle of the dust too, so matching the gas makes the drag zero. The convergence holds for every $\epsilon>0$ in this linear model; at $\epsilon=0$ transients never decay, so the zero-drag and late-time limits do not commute.
If instead the printed scalar equation is treated literally for finite $\epsilon$, its particular solution is
$$
\boxed{v'_{x,p}=u\left[\frac4{4+\epsilon^2}\cos\theta+\frac{2\epsilon}{4+\epsilon^2}\sin\theta\right].}
$$
It has amplitude $|u|/\sqrt{1+\epsilon^2/4}$ and phase offset $\arctan(\epsilon/2)$, not exact phase/amplitude agreement. Agreement in that truncated equation means only leading order as $\epsilon\to0$. This distinguishes the intended approximation from a false exact reading of the printed assertion.
There is a further order-of-limits condition: a real finite-wavelength gas wave has $\omega-\Omega\ne0$. Leading resonant entrainment over the damping time requires $|\omega-\Omega|\ll\epsilon\Omega$, or $(kc_s/\Omega)^2\ll\epsilon$. Otherwise solve the dust transfer equations at the actual $\omega$; detuning can reduce the dust response even though $|k|c_s/\Omega\ll1$.
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