= Solution
To first order in $F$, <time-dependent perturbation theory> replaces the state on the right-hand side by its unperturbed value $e^{-iE_0t/\hbar}|0\rangle$. The bound state is even, while the <position operator> $x$ is odd. The <parity> selection rule therefore gives
$$
\langle k,+|x|0\rangle=0,
$$
so $\dot b_k=0$ and, with no initial continuum component,
$$
\boxed{b_k(t)=0}.
$$
Projection onto the odd state gives
$$
\dot c_k(t)
=\frac{iF}{\hbar}\langle k,-|x|0\rangle
e^{i(E_k-E_0-\hbar\omega)t/\hbar}.
$$
Writing $\Omega_k=(E_k-E_0-\hbar\omega)/\hbar$ and integrating from zero to $t$ yields
$$
\boxed{
c_k(t)=\frac{iF}{\hbar}\langle k,-|x|0\rangle
e^{i\Omega_kt/2}
\frac{\sin(\Omega_kt/2)}{\Omega_k/2}}.
$$
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