= Solution
For fixed $\omega^2>0$ and weak <gas drag>, expand a wave root as $s=s_0+\epsilon s_1+O(\epsilon^2)$ with $s_0=\pm i\omega$. At order $\epsilon$, the <dispersion relation> yields
$$
(3s_0^2+\omega^2)s_1+2\Omega s_0^2+\Omega(\omega^2-\Omega^2)=0.
$$
Since $s_0^2=-\omega^2$, this reduces to $-2\omega^2s_1-\Omega(\omega^2+\Omega^2)=0$. The <weak-drag dust density waves> therefore have
$$
\boxed{s_\pm=\pm i\omega-\epsilon\frac{\Omega(\omega^2+\Omega^2)}{2\omega^2}+O(\epsilon^2).}
$$
The first-order real part is negative: the two density waves are \b[weakly damped], with their frequencies unchanged to first order. <Gas drag> removes their perturbation energy relative to the fixed gas flow. The expansion is an asymptotic result at fixed positive $\omega^2$; it is not uniform close to $\omega^2=0$, where the nominal damping can become comparable to the oscillation frequency and a different scaling is needed.
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