= Solution
Set $a=2\pi G\rho_0$. The axisymmetric <Poisson equation for Newtonian gravity> integrates to $(R\Phi_R)'=4\pi G\rho_0R$. Regularity on the axis fixes the integration constant, so $\Phi_R=aR$. Radial force balance in the <rotating fluid> gives
$$
p_R=\rho_0(R\Omega^2-\Phi_R)=\rho_0(k^2R^3-aR),\qquad
p=p_c+\rho_0\left(\frac{k^2R^4}{4}-\frac{aR^2}{2}\right).
$$
With the stated central <pressure>, $p_c=\rho_0a^2/(4k^2)$, this is a perfect square:
$$
\boxed{p(R)=\frac{\rho_0k^2}{4}(R^2-R_0^2)^2,\qquad R_0^2=\frac{a}{k^2}=\frac{2\pi G\rho_0}{k^2}.}
$$
The <free surface> lies at the first zero of <pressure>. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G\rho_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is $\Phi_R-R\Omega^2=R(a-k^2R^2)$, so \b[$\boxed{g_{\rm eff}(R_0)=0}$]. In particular $p_R(R_0)=0$. This is the <critically rotating self-gravitating cylinder>.
For a <normal mode> proportional to $e^{i\omega t+im\phi}$, the background <material derivative> becomes $i\sigma$, where $\sigma=\omega+m\Omega(R)$. Linearizing the centrifugal term gives $-2\Omega u_\phi$ in the radial equation. In the azimuthal equation the background <velocity> gradient and cylindrical geometric term together give $(2\Omega+R\Omega')u_R=3\Omega u_R$. Since the interior <mass density> is constant, write $W=p'/\rho_0+\Phi'$. The two momentum equations and <incompressible flow> condition are consequently
$$
\boxed{\begin{aligned}
i\sigma u_R-2\Omega u_\phi&=-W',\\
3\Omega u_R+i\sigma u_\phi&=-imW/R,\\
u_R'+u_R/R+im u_\phi/R&=0.
\end{aligned}}
$$
For $m\ne0$, the last two equations express the other amplitudes in terms of $u=u_R$:
$$
u_\phi=\frac{i}{m}(Ru'+u),\qquad W=\frac{iR}{m^2}\{3m\Omega u-\sigma(Ru'+u)\}.
$$
Differentiate the second expression using $R\Omega'=\Omega$ and $R\sigma'=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result is
$$
-\sigma R^2u''-3\sigma Ru'+[3m\Omega+(m^2-1)\sigma]u=0.
$$
Away from corotation this gives the required <ordinary differential equation>
$$
\boxed{u''+\frac3R u'+\frac{u}{R^2}\left(1-m^2-\frac{3m\Omega}{\sigma}\right)=0.}
$$
At corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial <velocity> would vanish by the <incompressible flow> condition; the requested dipolar calculation has $m=1$.
For $m=1$, use the amplitude normalization $u=A(\omega+kR)$, which remains useful when $\omega$ tends to zero. Substitution into the undivided equation gives zero identically. For nonzero $\omega$ this is the source's $1+kR/\omega$ after choosing $A=1/\omega$. The other amplitudes are
$$
\boxed{u_R=A(\omega+kR),\qquad u_\phi=iA(\omega+2kR),\qquad W=iAR[(kR)^2-\omega^2].}
$$
These formulas also satisfy the three original equations directly. By the regularity assumption in the question, this is the interior <velocity> family to which the <free surface> conditions must be applied.
The <gravitational potential of a displaced cylindrical interface> supplies an essential part of those conditions. For surface <fluid displacement> $\eta e^{im\phi}$, the bulk <mass density> perturbation vanishes but the displaced <mass density> step contributes $\rho_0\eta\delta(R-R_0)$. The interior and exterior <gravitational potential> amplitudes solve <Laplace's equation>, are proportional to $R^m$ and $R^{-m}$ for $m\geq1$, and are continuous at the surface. Integrating the <Poisson equation for Newtonian gravity> across the <mass density> step gives
$$
\Phi'_{R,\mathrm{out}}-\Phi'_{R,\mathrm{in}}=4\pi G\rho_0\eta.
$$
Since the two radial derivatives at the surface are $-m\Phi'_s/R_0$ and $m\Phi'_s/R_0$, respectively,
$$
\boxed{\Phi'_s=-\frac{aR_0}{m}\eta.}
$$
It would be incorrect to set the entire potential perturbation to zero merely because the bulk <mass density> is unchanged.
Let $\Omega_s=kR_0$, so $a=\Omega_s^2$. Zero <fluid Lagrangian perturbation> of surface <pressure> gives $p'_s+\eta p_R(R_0)=0$, hence $p'_s=0$. The surface momentum potential and kinematic condition for $m=1$ are therefore
$$
W_s=-\Omega_s^2R_0\eta,\qquad i(\omega+\Omega_s)\eta=A(\omega+\Omega_s).
$$
If $\omega+\Omega_s\ne0$, the kinematic condition fixes $\eta=-iA$. The same value follows if the radial <fluid Lagrangian displacement> $u_R/[i(\omega+kR)]=-iA$ is required to extend continuously to the <free surface>, including a possible surface-corotation limit. Comparing the gravitational condition with the interior value of $W_s$ then gives
$$
iAR_0(\Omega_s^2-\omega^2)=iAR_0\Omega_s^2,\qquad \boxed{\omega^2=0}
$$
for a nontrivial perturbation. These <dipolar perturbations of a critically rotating cylinder> describe a displacement of the whole cylinder's axis. Translating an isolated self-gravitating system gives another equilibrium, so there is no external restoring force and the translation is neutral. A uniform drift of the centre also follows from invariance under a <Galilean transformation>, explaining the double zero of the frequency equation. The singular $1/\omega$ in the originally chosen <velocity> normalization does not obstruct the finite-amplitude limiting solution.
There is an admissibility qualification to the frequency conclusion as printed. Regular interior <fluid Eulerian perturbation> of <velocity> alone does not justify cancelling the surface factor $\omega+\Omega_s$. Eliminating $\eta$ without that cancellation instead gives
$$
A\omega^2(\omega+\Omega_s)=0.
$$
The additional surface-corotation case $\omega=-\Omega_s$ formally satisfies every displayed <velocity> equation and the usual linearized surface conditions with
$$
\eta=0,\qquad \Phi'=0,\qquad u_R=A(kR-\Omega_s),\qquad
u_\phi=iA(2kR-\Omega_s),\qquad p'=\rho_0iAR[(kR)^2-\Omega_s^2].
$$
Here $u_R(R_0)=p'(R_0)=0$, and the <velocity> is finite on the axis. However, its interior radial <fluid Lagrangian displacement> tends to $-iA$, whereas the undisplaced surface has $\eta=0$. Thus a regular single-frequency <fluid Lagrangian displacement> continuous up to the surface excludes this degenerate branch. \b[The intended $\omega^2=0$ conclusion holds with that regular-displacement condition, or with surface corotation excluded; the printed <velocity> regularity assumption by itself leaves this formal exception.]
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