= Solution
At first order in <time-dependent perturbation theory> set $c_1\simeq1$ and use the <parity> result $V_{22}=0$. Decomposing the cosine into its two exponentials gives
$$
c_2(t)=-\frac{i\Omega}{2}\int_0^t
[e^{i(\omega_0-\omega)t'}+e^{i(\omega_0+\omega)t'}]dt'.
$$
Both exponentials contribute to the exact first-order result:
$$
\boxed{c_2(t)=-\frac\Omega2\left[
\frac{e^{i(\omega_0-\omega)t}-1}{\omega_0-\omega}
+\frac{e^{i(\omega_0+\omega)t}-1}{\omega_0+\omega}\right].}
$$
Each apparent zero denominator has a continuous limit. In particular the near-resonant fraction tends to $it$ as $\omega\to\omega_0$, while the counter-rotating fraction remains bounded on the optical timescale. One may equivalently write the expression as
$$
c_2(t)=-\frac{i\Omega t}{2}\sum_{\epsilon=\pm1}
e^{i(\omega_0+\epsilon\omega)t/2}
\operatorname{sinc}[(\omega_0+\epsilon\omega)t/2],
\qquad\operatorname{sinc}u=\frac{\sin u}{u}.
$$
This form makes $c_2(0)=0$ explicit and avoids cancellation at resonance. At extremely short time both <sinc functions> tend to one, so $c_2\simeq-i\Omega t$, agreeing with the initial derivative of the original cosine-driven equation. No <rotating-wave approximation> has yet been made. The separate assumption of weak excitation remains necessary: the first-order expression must not be extended until it predicts <probabilities> comparable to or larger than one.
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