= Solution
Normalize the incident intensity to one. Since $|K^0\rangle=(|K_S\rangle+|K_L\rangle)/\sqrt2$, the initial <probability amplitudes> are $A_S(0)=A_L(0)=1/\sqrt2$. Define $g_j=e^{-im_jt}e^{-t/(2\tau_j)}$. The flavor <probability amplitudes> are $(g_S+g_L)/2$ and $(g_S-g_L)/2$. Their absolute squares are
$$
\boxed{I_{K^0}(t)=\frac14\left[e^{-t/\tau_S}+e^{-t/\tau_L}
+2e^{-\frac t2(1/\tau_S+1/\tau_L)}\cos(\Delta m\,t)\right],}
$$
$$
\boxed{I_{\bar K^0}(t)=\frac14\left[e^{-t/\tau_S}+e^{-t/\tau_L}
-2e^{-\frac t2(1/\tau_S+1/\tau_L)}\cos(\Delta m\,t)\right],}
$$
where $\Delta m=m_L-m_S$ in <natural units>. Thus $A=B=1/4$ for both, $\phi=\Delta m\,t$, and
$$
\boxed{C_{K^0}(t)=\tfrac12e^{-\frac t2(1/\tau_S+1/\tau_L)},\qquad
C_{\bar K^0}(t)=-\tfrac12e^{-\frac t2(1/\tau_S+1/\tau_L)}.}
$$
Multiplying all three coefficients by the incident intensity gives any other normalization. Crucially, \b[$C$ is time dependent]. The printed template cannot hold with all of $A,B,C$ constant. For example, a nonzero constant $C$ would leave an undamped oscillation at arbitrarily late times, even after both unstable components have decayed.
In seconds, $\phi=\Delta m\,c^2t/\hbar$. With $x=t/\tau_S$, the supplied <mass> difference and <mean lifetimes> give
$$
r_\tau=\tau_S/\tau_L=0.00172934,\qquad
\phi=0.474975x,
$$
$$
I_{K^0,\bar K^0}(x)=\frac14\left[e^{-x}+e^{-r_\tau x}
\mathbin{\pm}2e^{-(1+r_\tau)x/2}\cos(0.474975x)\right].
$$
Initially the intensities are one and zero. Their sum is $\tfrac12(e^{-x}+e^{-r_\tau x})$, so these are unconditioned survival intensities, not <probabilities> conditioned on survival. The curves first meet at $x=\pi/(2\times0.474975)\simeq3.307$. Thereafter the <neutral antikaon> component briefly exceeds the $K^0$ component, before the <interference> dies away and both approach the surviving long-lived intensity $e^{-r_\tau x}/4$.
\Image[/past-exam-of-the-nuclear-and-particle-physics-course-of-the-university-of-oxford/2022-kaon-intensities.png]
{title=Neutral-kaon flavor intensities for an initially pure neutral-kaon beam, versus proper time}
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