= Solution
Let $\Sigma_0$ be the uniform background <surface density>, $\mathbf u_0=-Sx\mathbf e_y$, and write $\Sigma=\Sigma_0+\Sigma'$, $\mathbf u=\mathbf u_0+\mathbf v$, $\Phi_d=\Phi_{d0}+\Phi_d'$. The background sheet has no horizontal gravitational <gradient>; its vertical field is irrelevant to the stated planar equations. Put $D_0=\partial_t-Sx\partial_y$. Keeping first-order terms gives
$$
\boxed{D_0\Sigma'+\Sigma_0(\partial_xv_x+\partial_yv_y)=0,}
$$
$$
\boxed{D_0v_x-2\Omega v_y=-\partial_x\left(c_s^2\frac{\Sigma'}{\Sigma_0}+\Phi_{d,m}'\right),\qquad D_0v_y+(2\Omega-S)v_x=-\partial_y\left(c_s^2\frac{\Sigma'}{\Sigma_0}+\Phi_{d,m}'\right),}
$$
$$
\boxed{\nabla^2\Phi_d'=4\pi G\Sigma'\delta(z),\qquad\Phi_{d,m}'=\Phi_d'(x,y,0,t).}
$$
The $-Sv_x$ contribution comes from $\mathbf v\cdot\nabla\mathbf u_0$, while the <Coriolis acceleration> supplies the remaining coupling. The prescribed tidal potential has no perturbation, and its background force already balances the background Coriolis term. The isothermal <pressure> linearizes to $c_s^2\nabla\Sigma'/\Sigma_0$.
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