= Solution
The <pressureless dust fluid in a shearing sheet> has no <pressure> restoration and is driven by <drag force> toward the prescribed gas velocity; gas backreaction is omitted in the given model. After linearizing about the common background shear, axisymmetry gives
$$
(\partial_t+\epsilon\Omega)v'_x-2\Omega v'_y=\epsilon\Omega u'_x,\qquad (\partial_t+\epsilon\Omega)v'_y+\frac\Omega2v'_x=\epsilon\Omega u'_y.
$$
Apply $D_\epsilon=\partial_t+\epsilon\Omega$ to the first equation and use the second to eliminate $v'_y$. The exact linear scalar equation is
$$
\bigl[\partial_t^2+2\epsilon\Omega\partial_t+(1+\epsilon^2)\Omega^2\bigr]v'_x=\epsilon\Omega\partial_tu'_x+\epsilon^2\Omega^2u'_x+2\epsilon\Omega^2u'_y.
$$
For the supplied leading-order epicycle, set $\theta=kx-\Omega t$. Since $\partial_tu'_x=\Omega u\sin\theta$ and $u'_y=u\sin\theta/2$, this becomes
$$
\boxed{\partial_t^2v'_x+2\epsilon\Omega\partial_tv'_x+(1+\epsilon^2)\Omega^2v'_x=2\epsilon\Omega^2u\sin\theta+\epsilon^2\Omega^2u\cos\theta.}
$$
The final cosine term is missing from the printed equation. Dropping it is justified at leading order in $0<\epsilon\ll1$, recovering the intended forcing, but the printed equation retains an order-$\epsilon^2$ frequency term on the left and is not an exact derivation at that order. A direct countercheck is $\mathbf v'=\mathbf u'$: the drag vanishes and the pair solves both dust equations exactly, while $v'_x=u\cos\theta$ leaves residual $\epsilon^2\Omega^2u\cos\theta$ in the printed scalar equation. This is the <forced dust epicycle with aerodynamic drag>; its damping time is $(\epsilon\Omega)^{-1}$.
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