 Cubic axisymmetric inertial mode of a rigid sphere

For an axisymmetric normal mode in a rigid rotating sphere, substitute $W=z(A{R}^2+B{z}^2+C)$ into the pressure equation and its impermeability condition. With $\beta =1-4{\Omega }^2/{\sigma }^2$, the interior equation gives $2A+3\beta B=0$. The boundary polynomial gives $3\beta B=(2+\beta )A$ and $\beta C=-(2+\beta )A{r}_0^2$. A nontrivial oscillatory solution has $\beta =-4$, hence the displayed frequency and $B=A/6$, $C=-A{r}_0^2/2$. It satisfies both equations identically.