Solving the logistic differential equation

Let \(K\) represent the carrying capacity for a particular organism in a given environment, and let \(r\) be a real number that represents the growth rate. The function \(P(t)\) represents the population of this organism as a function of time \(t\). The logistic differential equation is:

\[ \frac{dP}{dt} =rP \left( 1 - \frac{P}{K} \right) \]

To solve this let's first start by rewriting the equation:

\[ \frac{dP}{dt} = \frac{rP(K-P)}{K} \]

Rearranging:

\[ \frac{K}{P(K-P)} \ dP = r \ dt \]

Now we can take the integral of both sides:

\[ \int \frac{K}{P(K-P)} \ dP = \int r \ dt \]

We can use partial fraction decomposition on the left-hand side:

\[ \int \frac{1}{P} + \frac{1}{K-P} \ dP = \int r \ dt \]

Evaluating:

\[\begin{aligned} \ln|P| - \ln|K-P| &= rt + C \\ \ln \left| \frac{P}{K-P} \right| &=rt+C\end{aligned} \]

Now exponentiate both sides:

\[\begin{aligned} \left| \frac{P}{K-P} \right| &= e^{rt+C} = e^{rt}e^C \\ &= e^{rt}D\end{aligned} \]

Since \(P > 0\) and \(K \gt P\), then \(\frac{P}{K-P} > 0\):

\[ \frac{P}{K-P} = e^{rt}D \]

If we make \(P\) the subject:

\[\begin{aligned} P &= DKe^{rt} - DPe^{rt} \\ P (1 + De^{rt}) &= DKe^{rt} \\ P &= \frac{DKe^{rt}}{1 + De^{rt}} \end{aligned} \]

To determine the value of \(D\), we can set \(t = 0\). Let \(P_0\) be the initial population:

\[\begin{aligned} \frac{P_0}{K-P_0} &= e^{r(0)}D \\ \frac{P_0}{K-P_0} &= D \end{aligned}\]

Substituting the expression for \(D\) into the equation for \(P\):

\[P = \frac{\frac{P_0}{K-P_0} Ke^{rt}}{1 + \frac{P_0}{K-P_0} e^{rt}} \]

Multiply the numerator and denominator by \((K-P_0)\):

\[P(t) = \frac{P_0 Ke^{rt}}{(K-P_0) +P_0 e^{rt}} \]

Solving The Logistic Differential Equation