\end It is easy to verify that \begin P(t) = \frac \end is the solution to our initial value problem.

If we differentiate \(P\text\) we will obtain the righthand side of the **differential** **equation**, \begin \frac & = \frac \left(\frac\right)\ & = 1000 k \frac\ & = k \frac \cdot \frac\ & = k \frac \cdot \frac\ & = k \left( 1 - \frac\right) \frac\ & = k \left( 1 - \frac \right) P, \end and certainly \(P(0) = 100\text\) In addition, if we know that the population is 200 fish after one year, then \begin 200 = P(1) = \frac \end and we can determine that \begin k = \ln\left( \frac\right) \approx 0.8109.

If we also assume that the population has a constant death rate, the change in the population \(\Delta P\) during a small time interval \(\Delta t\) will be \begin \Delta P \approx k_ P(t) \Delta t - k_ P(t) \Delta t, \end where \(k_\) is the fraction of the population having offspring during the interval and \(k_\) is the fraction of the population that dies during the interval.

Equivalently, we can write \begin \frac \approx k P(t), \end where \(k = k_ - k_\text\) Since the derivative of \(P\) is \begin \frac = \lim_ \frac, \end the rate of change of the population is proportional to the size of the population, or \begin \frac = k P.\label\tag \end The **equation** \begin \frac = k P \end is one of the simplest **differential** **equations** that we will consider.

\end Thus, the Suppose we have a pond will support 1000 fish, and the initial population is 100 fish.

\end In addition, if we know the value of \(P(t)\text\) say when \(t = 0\text\) we can also determine the value of \(C\text\) For example, if the population at the time \(t = 0\) is \(P(0) = P_0\text\) then \begin P_0 = P(0) = Ce^ = C \end or \(P(t) = P_0 e^\text\) The **differential** **equation** \begin P'(t) & = k P(t),\ P(0) & = P_0 \end is an example of an .

Consequently, \begin k = \ln 1.03 \approx 0.0296 \end and the solution to our initial value problem is \begin P(t) = 1000e^.

\end Of course, it is important to realize that this is only a model.

As an example, suppose that \(P(t)\) is a population of a colony of bacteria at time \(t\text\) whose initial population is 1000 at \(t = 0\text\) where time is measured in hours.

Then \begin 1000 = P(0) = C e^0 = C, \end and our solution becomes \(P(t) = 1000e^\text\) If the population grows at three percent per hour, then \begin 1030 = P(1) = 1000 e^k, \end after one hour.

\end In addition, if we know the value of \(P(t)\text\) say when \(t = 0\text\) we can also determine the value of \(C\text\) For example, if the population at the time \(t = 0\) is \(P(0) = P_0\text\) then \begin P_0 = P(0) = Ce^ = C \end or \(P(t) = P_0 e^\text\) The **differential** **equation** \begin P'(t) & = k P(t),\ P(0) & = P_0 \end is an example of an .Consequently, \begin k = \ln 1.03 \approx 0.0296 \end and the solution to our initial value problem is \begin P(t) = 1000e^.\end Of course, it is important to realize that this is only a model.As an example, suppose that \(P(t)\) is a population of a colony of bacteria at time \(t\text\) whose initial population is 1000 at \(t = 0\text\) where time is measured in hours.Then \begin 1000 = P(0) = C e^0 = C, \end and our solution becomes \(P(t) = 1000e^\text\) If the population grows at three percent per hour, then \begin 1030 = P(1) = 1000 e^k, \end after one hour.In addition, the theory of the subject has broad and important implications.