= Solution
Use perturbations $\mathbf v,\mathbf b,p,\vartheta$ for <velocity>, <magnetic field>, <magnetohydrodynamic total pressure> and <buoyancy displacement variable>. Set
$$
D_0=\partial_t-\frac32\Omega x\partial_y.
$$
This is advection by the background <Keplerian shearing sheet>. Subtracting the equilibrium before retaining first-order terms gives \b[the nine linearized equations]:
$$
\begin{aligned}
D_0v_x-2\Omega v_y&=-\rho_0^{-1}\partial_xp-N^2\vartheta+\frac{B_0}{4\pi\rho_0}\partial_zb_x,\\
D_0v_y+\frac12\Omega v_x&=-\rho_0^{-1}\partial_yp+\frac{B_0}{4\pi\rho_0}\partial_zb_y,\\
D_0v_z&=-\rho_0^{-1}\partial_zp+\frac{B_0}{4\pi\rho_0}\partial_zb_z,\\
D_0b_x&=B_0\partial_zv_x,\\
D_0b_y&=-\frac32\Omega b_x+B_0\partial_zv_y,\\
D_0b_z&=B_0\partial_zv_z,\\
D_0\vartheta&=v_x,\\
\partial_xv_x+\partial_yv_y+\partial_zv_z&=0,\\
\partial_xb_x+\partial_yb_y+\partial_zb_z&=0.
\end{aligned}
$$
The azimuthal coefficient $\Omega/2$ combines the <Coriolis acceleration> with perturbation advection of the background shear. The $-3\Omega b_x/2$ term is the winding of radial <magnetic field> by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. <Magnetic pressure> is already included in $p$, leaving only linear <magnetic tension>. In this normalization $\vartheta$ has dimensions of length because $D_0\vartheta=v_x$.
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