= Solution
Write the proposed background as $\mathbf U=U(x,z)\mathbf e_y$, $U=-3\Omega x/2+q\Omega z$. It satisfies <incompressibility>, and $\mathbf U\cdot\nabla\mathbf U=U\partial_y\mathbf U=0$. A constant <pressure> has zero <gradient>. The <Coriolis acceleration> is $-2\Omega\mathbf e_z\times\mathbf U=2\Omega U\mathbf e_x=-(3x-2qz)\Omega^2\mathbf e_x$, which exactly cancels the prescribed body force. Thus it is a <steady state>.
\b[The background is ordinary Keplerian radial shear together with vertical shear:] $\partial_xU=-3\Omega/2$ and $\partial_zU=q\Omega$. At a given radius, different heights have different azimuthal velocities. The small $q$ in this question measures vertical shear; it is distinct from the <orbital shear parameter>, whose radial value here remains $3/2$. The <vertical shear instability> draws on this height-dependent differential rotation.
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