Set $\tau=t-t_*$. The <Logistic solution of the SI model> becomes
$$
\Lambda(\tau)=\frac{N\beta e^{-\beta\tau}}
{(1+e^{-\beta\tau})^2}
=\frac{N\beta}{4\cosh^2(\beta\tau/2)}.
$$
It is a symmetric, unimodal bell-shaped curve about the incidence peak $\tau=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 3 / e / Solution

{c}

If $S(0)=N\theta/(1+\theta)$ and $I(0)=N/(1+\theta)$, then
$$
I(t)=\frac{N}{1+\theta e^{-\beta t}},
\qquad
\dot I(t)=\frac{N\beta\theta e^{-\beta t}}{(1+\theta e^{-\beta t})^2}.
$$
After centering time at the incidence peak $t_*=\beta^{-1}\log\theta$, the incidence curve is $N\beta/[4\cosh^2\{\beta(t-t_*)/2\}]$.


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 Incoming links: Logistic solution of the SI model