To see how the logistic model works, let us try to adjust our model of exponential growth to account for the limited resources of the pond. We will make the following assumptions.
If the population of trout is small and the pond is large with abundant resources, the rate of growth will be approximately exponential,
\begin{equation*}
\frac{dP}{dt} \approx kP.
\end{equation*}
If \(N\) is the maximum population of trout that the pond can support, then any population larger than \(N\) will decrease. In other words,
\begin{equation*}
\frac{dP}{dt} \lt 0
\end{equation*}
for \(P \gt N\text{.}\) We say that \(N\) is the carrying capacity for the population.
O ur assumptions suggest that we might try an equation of the form
\begin{equation*}
\frac{dP}{dt} = k f(P) P,
\end{equation*}
where \(f(P)\) is a function of \(P\) that is close to 1 if the population is small but negative if the population is greater than \(N\text{.}\) The simplest function satisfying these conditions is
\begin{equation*}
f(P) = \left( 1 - \frac{P}{N} \right).
\end{equation*}
Thus, the logistic population model is given by the differential equation
\begin{equation*}
\frac{dP}{dt} = k \left( 1 - \frac{P}{N} \right) P.
\end{equation*}