Houjun Liu

logistic equation


\begin{equation} P’ = 2P(100-P) \end{equation}

for a motivation, see petri dish.


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}


  • as \(t \to -\infty\), we have \(p \to 0\)
  • as \(t \to +\infty\), we have \(p \to 100\)