Example 1.1.2.
Suppose we have a pond that will support 1000 fish, and the initial population is 100 fish. In order to determine the number of fish in the lake at any time \(t\text{,}\) we must find a solution to the initial value problem
\begin{align*}
\frac{dP}{dt} & = k \left( 1 - \frac{P}{1000} \right) P\\
P(0) & = 100.
\end{align*}
It is easy to verify that \(P(t) = 1000/(9e^{-kt} + 1)\) is the solution to our initial value problem. Certainly \(P(0) = 100\text{,}\) and if we differentiate \(P\text{,}\) we will obtain the righthand side of the differential equation,
\begin{align*}
\frac{dP}{dt} & = \frac{d}{dt} \left(\frac{1000}{9e^{-kt} + 1}\right)\\
& = 1000 k \frac{9e^{-kt}}{(9e^{-kt} + 1)^2}\\
& = k \frac{9e^{-kt}}{(9e^{-kt} + 1)} \cdot \frac{1000}{9e^{-kt} + 1}\\
& = k \frac{(9e^{-kt} + 1) - 1}{(9e^{-kt} + 1)} \cdot \frac{1000}{9e^{-kt} + 1}\\
& = k \left( 1 - \frac{1000}{1000(9e^{-kt} + 1)}\right) \frac{1000}{9e^{-kt} + 1}\\
& = k \left( 1 - \frac{P}{1000} \right) P.
\end{align*}
In addition, if we know that the population is 200 fish after one year, then
\begin{equation*}
200 = P(1) = \frac{1000}{9e^{-k} + 1},
\end{equation*}
and we can determine that
\begin{equation*}
k = \ln\left( \frac{9}{4}\right) \approx 0.8109.
\end{equation*}
Consequently, the solution to our intial-value problem is
\begin{equation*}
P(t) = \frac{1000}{9e^{-0.8109 t} + 1}.
\end{equation*}
The graph of our solution certainly fits the situation that we are modeling (Figure 1.1.3). We will learn how to solve initial value problems such as the one described here in Section 1.2. For the time being, we will be satisfied with being able to verify the fact that we have a solution.
a curve that first increases steeply and then begins to level off just below one thousand