= Solution
First distinguish concentration conventions. If $C$ is cell number density, the excess mixture density is $v\Delta\rho C$. If $\varphi=vC$ is cell volume fraction, it is $\Delta\rho\varphi$. \b[The printed definition of $R$ is dimensionless for number density, whereas the prose calls $\hat n_0$ a volume fraction.] Use $C_0$ for mean number density and $\varphi_0=vC_0$ for mean volume fraction. Then the consistent <Rayleigh number> is
$$
\boxed{R=\frac{gC_0v\Delta\rho H^3}{\rho\nu D}=\frac{g\varphi_0\Delta\rho H^3}{\rho\nu D}.}
$$
The dimensionless concentration $n=C/C_0=\varphi/\varphi_0$ is the same in both conventions. The extra factor $v$ must be removed if the source symbol is literally a volume fraction.
Remove the hydrostatic basic pressure and use a dilute <Boussinesq approximation>. With $\widehat C_e=\widehat C-C_0\bar n$ and perturbation fluid <velocity> $\widehat{\mathbf u}$, the linear incompressible <Newtonian fluid> equations are
$$
\widehat\nabla\cdot\widehat{\mathbf u}=0,\qquad
\partial_{\hat t}\widehat{\mathbf u}=-\frac1\rho\widehat\nabla\widehat P_e+\nu\widehat\Delta\widehat{\mathbf u}-\frac{gv\Delta\rho}{\rho}\widehat C_e\mathbf e_3.
$$
The <cell conservation equation> linearizes to
$$
\partial_{\hat t}\widehat C_e+\widehat w\,C_0\frac{d\bar n}{d\hat z}+V_s\partial_{\hat z}\widehat C_e+V_sC_0\widehat\nabla_\perp\cdot(\bar n\mathbf p_\perp)=D\widehat\Delta\widehat C_e.
$$
At this order the vertical orientation perturbation is zero, because $|\mathbf p|=1$. The instantaneous-orientation approximation from the introductory calculation gives
$$
\mathbf p_\perp=\frac B2(\widehat{\boldsymbol\omega}\times\mathbf e_3)_\perp.
$$
Incompressibility implies $\widehat\nabla_\perp\cdot(\widehat{\boldsymbol\omega}\times\mathbf e_3)_\perp=-\widehat\Delta\widehat w$. This identity fixes the sign of the gyrotactic term.
Apply the length, viscous-time, <velocity>, concentration and pressure scales in the paper. Writing $n_e=n-\bar n$, the dimensionless equations are
$$
\begin{aligned}
\nabla\cdot\mathbf u&=0,\\
\partial_t\mathbf u&=-\nabla P_e+\Delta\mathbf u-Rn_e\mathbf e_3,\\
S\partial_tn_e&=\Delta n_e-h\partial_zn_e-\bar n'w+G\bar n\Delta w,
\end{aligned}
$$
where $S=\nu/D$ and $G=V_sB/(2H)$. The factor $S$ appears in the cell equation because its diffusive time differs from the viscous time used for $t$.
For a <normal mode> $[n_e,w]=[N,W]e^{\sigma t+i\kappa x}$, set $\mathcal L=d^2/dz^2-\kappa^2$. Taking the divergence of the momentum equation gives $\mathcal LP_e=-RN'$. Applying $\mathcal L$ to its vertical component eliminates pressure and yields
$$
\boxed{(\mathcal L-\sigma)\mathcal LW=-R\kappa^2N.}
$$
The concentration equation becomes
$$
\boxed{\left(\frac{d^2}{dz^2}-h\frac d{dz}-\kappa^2-S\sigma\right)N=\bar n'W-G\bar n\left(\frac{d^2}{dz^2}-\kappa^2\right)W.}
$$
These reproduce the requested modal equations, with the physically consistent sign of the orientation balance and concentration convention made explicit.
At both plates, the <no-slip boundary condition> gives $w=0$ and horizontal perturbation <velocity> zero. For a nonzero horizontal <wavenumber>, incompressibility then gives $W'=0$. The normal cell flux is $\bar n w+hN-N'$; the orientation correction has no vertical component at this order. Impermeability therefore gives
$$
\boxed{W=W'=0,\qquad N'-hN=0\quad\text{at }z=-1,0.}
$$
If orientation lag is retained rather than the given instantaneous balance, $\mathbf p_\perp$ acquires the relaxation response $(1+B\partial_{\hat t})^{-1}$ and the modal gyrotactic term changes accordingly. The displayed equations are the quasistatic-orientation model requested here.
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