= Solution
The steady radial equation first gives the base-pressure gradient
$$
\frac{dp_0}{dr}=r\Omega^2.
$$
For the stated <normal mode>, put
$$
D(r)=\sigma+im\Omega(r).
$$
Retaining terms linear in the disturbance in the <Euler equations for an inviscid fluid> gives
$$
\boxed{Du'-2\Omega v'+\frac{dp'}{dr}=0},
$$
$$
\boxed{Dv'+(2\Omega+r\Omega')u'
+\frac{im}{r}p'=0},
$$
$$
\boxed{Dw'+ikp'=0},
$$
together with <incompressible flow>
$$
\boxed{\frac1r\frac d{dr}(ru')
+\frac{im}{r}v'+ikw'=0}.
$$
The $-2\Omega v'$ and $(2\Omega+r\Omega')u'$ terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.
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