= Solution
For $\Psi=f(R)\sin^2\theta$, the angular differentiation gives
$$
\mathcal D^2\Psi=(f''-2f/R^2)\sin^2\theta.
$$
Trying $f=R^m$ in the fourth-order equation gives the indicial product
$$
[m(m-1)-2][(m-2)(m-3)-2]=0,
$$
whose four distinct roots are $-1,1,2,4$. This proves that the stated four-term expression is the general separated solution of this angular type.
Put $z=b/a$ and $F(s)=A/s+Bs+Cs^2+Ds^4$. The velocity formulas become $u_R=2U F(s)\cos\theta/s^2$ and $u_\theta=-U F'(s)\sin\theta/s$. The four boundary conditions therefore give
$$
\boxed{\begin{aligned}
A+B+C+D&=\tfrac12,\\
-A+B+2C+4D&=0,\\
A/z+Bz+Cz^2+Dz^4&=0,\\
-A/z^2+B+2Cz+4Dz^3&=0.
\end{aligned}}
$$
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