Consider:
\begin{equation} P’ = 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 \left(\frac{1}{p} + \frac{1}{100-p}\right)\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