= Solution
Insert the stated expansion into the <Time-dependent Schrödinger equation>
$$
i\hbar\frac d{dt}|\psi(t)\rangle
=(H_0+v(t))|\psi(t)\rangle.
$$
When the derivative acts on each free phase, its energy factor is cancelled by the action of $H_0$ on the corresponding <eigenstate>. The remaining terms obey
$$
i\hbar\left[
\dot a(t)e^{-iE_0t/\hbar}|0\rangle
+\int_0^\infty e^{-iE_kt/\hbar}
\bigl(\dot b_k(t)|k,+\rangle+\dot c_k(t)|k,-\rangle\bigr)\,dk
\right]
=-F e^{-i\omega t}x|\psi(t)\rangle.
$$
Division by $i\hbar$ gives
$$
\boxed{
\dot a(t)e^{-iE_0t/\hbar}|0\rangle
+\int_0^\infty e^{-iE_kt/\hbar}
\bigl(\dot b_k|k,+\rangle+\dot c_k|k,-\rangle\bigr)\,dk
=\frac{iF}{\hbar}e^{-i\omega t}x|\psi(t)\rangle}.
$$
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