Suggest edit — logistic equation

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title: "logistic equation"
source: https://www.jemoka.com/posts/kbhlogistic_equations/
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Consider:
\begin{equation} P&rsquo; = 2P(100-P) \end{equation}
for a motivation, see petri dish.
Solution Assuming \(P\) never reaches 100
\begin{equation} \int \frac{\dd{P}}{P(100-P)} \dd{P}= \int 2 \dd{t} \end{equation}
Partial fractions time:
\begin{equation} \frac{1}{100} \int \qty(\frac{1}{p} + \frac{1}{100-p})\dd{P} = \frac{1}{100} \ln |p| - \ln |100-p| = 2t+C \end{equation}
Remember now log laws:
\begin{equation} \frac{1}{100} \ln \left| \frac{p}{100-p} \right| = 2t+C \end{equation}
And finally, we obtain:
\begin{equation} \qty | \frac{p}{100-p} | = e^{200t + C} \end{equation}
We can get rid of the absolute value by reshaping the fraction:
\begin{equation} \frac{p}{100-p} = ke^{200t} \end{equation}
Finally, we solve for \(p\):
\begin{equation} p(t) = \frac{100k e^{200t}}{1+ke^{200t}} = \frac{100k}{e^{-200t}+k} \end{equation}
Note!
as \(t \to -\infty\), we have \(p \to 0\) as \(t \to +\infty\), we have \(p \to 100\)