= Solution
The far field is proportional to $rP_1(\cos\theta)$, since $P_1(\cos\theta)=\cos\theta$. After matching the growing terms, write the exterior solution as
$$
U=v_0rP_1(\cos\theta)+\sum_{n=0}^\infty B_nr^{-n-1}P_n(\cos\theta).
$$
An added constant has here been excluded by interpreting the far-field condition as $U-v_0r\cos\theta\to0$. Differentiate and impose the radial <Neumann boundary condition>:
$$
0=v_0P_1(\cos\theta)-\sum_{n=0}^\infty(n+1)B_nr_0^{-n-2}P_n(\cos\theta).
$$
The <Orthogonality of Legendre polynomials> makes each coefficient vanish separately. Thus $B_n=0$ for $n\ne1$, and $2B_1/r_0^3=v_0$. Consequently
$$
\boxed{U(r,\theta)=v_0\left(r+\frac{r_0^3}{2r^2}\right)\cos\theta,\qquad r\geq r_0.}
$$
Indeed $\partial_rU=v_0(1-r_0^3/r^3)\cos\theta$, which vanishes at the boundary, while the correction decays at infinity. If the far-field condition is intended to specify only the leading growing term, an arbitrary additive constant is also allowed; its gradient has no effect on the <Neumann boundary condition>.
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