= Solution
Differentiate $\dot c_2=-i\gamma e^{2i\Omega t}c_1$ and substitute the equation for $\dot c_1$. The result is
$$
\boxed{\ddot c_2-2i\Omega\dot c_2+\gamma^2c_2=0.}
$$
Putting $c_2=e^{i\Omega t}u$ removes the first <derivative> and gives $\ddot u+\kappa^2u=0$, where $\kappa=\sqrt{\Omega^2+\gamma^2}$. For $\kappa>0$ the general solution is
$$
c_2=e^{i\Omega t}(A\cos\kappa t+B\sin\kappa t).
$$
The initial <spinor> fixes $c_1(0)=1,c_2(0)=0$, and the \b[first-order coupled equation] fixes the second initial datum $\dot c_2(0)=-i\gamma$. Thus $A=0$, $B=-i\gamma/\kappa$, and
$$
\boxed{c_2(t)=-i\frac\gamma\kappa e^{i\Omega t}\sin\kappa t,\qquad c_1(t)=e^{-i\Omega t}\left(\cos\kappa t+i\frac\Omega\kappa\sin\kappa t\right).}
$$
Their squared magnitudes sum to one. If both $\Omega$ and $\gamma$ vanish, the initial amplitudes simply remain $(1,0)$; the formulas have that continuous limit.
\b[Source convention correction.] For positive $\gamma$, the printed expression for $c_2$ lacks the factor $-i$ required by the stated <basis> and initial <vector>. A concrete test is its <derivative> at $t=0$: the printed expression gives $+\gamma$, whereas the <Schrödinger equation> gives $-i\gamma$. Multiplying only $c_2$ by a <phase> changes the relative <quantum phase>. One may change the overall <phase> of the entire <spinor>, but must then transform $c_1$ and its initial amplitude as well. The printed modulus nevertheless gives the correct <transition probability>.
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