Question

Solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{\mathrm{o}}\) the initial population. $$r=0.4, K=5500, P_{0}=100$$

Short answer

Answer: The population function that represents the logistic growth model for this problem is: $$ P(t) = \frac{5500}{1 + \left(\frac{5500}{100} - 1\right)e^{-0.4t}} $$

Step-by-step solution

Identify the logistic equation

We already have the logistic equation in terms of the problem statement: $$ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) $$ with given values, \(r=0.4\), \(K=5500\), \(P_{0}=100\). Step 2: Solve the differential equation

Separation of variables

We'll start by separating variables in the logistic equation. To do this, we'll divide both sides by \(P\left(1 - \frac{P}{K}\right)\) and multiply both sides by \(dt\): $$ \frac{dP}{P\left(1 - \frac{P}{K}\right)} = r \, dt $$ Step 3: Integrate both sides

Integrate

Integrate both sides of the equation with respect to their respective variables: $$ \int \frac{dP}{P\left(1 - \frac{P}{K}\right)} = \int r \, dt $$ Step 4: Use partial fraction decomposition

Partial fraction decomposition

To integrate the left-hand side, we need to perform partial fraction decomposition: $$ \frac{1}{P\left(1 - \frac{P}{K}\right)} = \frac{A}{P} + \frac{B}{1 - \frac{P}{K}} $$ By clearing the denominator, we find that \(A = \frac{1}{K}\) and \(B=\frac{1}{K}\). So the left-hand side integral becomes: $$ \int \left(\frac{1}{K}\frac{dP}{P} + \frac{1}{K}\frac{dP}{1 - \frac{P}{K}}\right) $$ Step 5: Integrate the left-hand side

Integration of the left-hand side

Now, we can integrate the left-hand side: $$ \int \left(\frac{1}{K}\frac{dP}{P} + \frac{1}{K}\frac{dP}{1 - \frac{P}{K}}\right) = \frac{1}{K} \int \frac{dP}{P} + \frac{1}{K} \int \frac{dP}{1 - \frac{P}{K}} $$ Applying the integration, we get: $$ \frac{1}{K} \ln|P| - \frac{1}{K} \ln|1 - \frac{P}{K}| = rt + C $$ Step 6: Solve for the particular solution

Particular solution

Now we solve for the particular solution given the initial condition \(P_{0}=100\): $$ \frac{1}{K} \ln\left|\frac{P_{0}}{K - P_{0}}\right| = C $$ Insert the known values to find \(C\): $$ \frac{1}{5500} \ln\left|\frac{100}{5500 - 100}\right| = C $$ Step 7: Express the population function

Population function

Now, we can express \(P(t)\) in terms of the constants found above: $$ P(t) = \frac{K}{1 + \left(\frac{K}{P_{0}} - 1\right)e^{-rt}} $$ Insert the known values to complete the population function: $$ P(t) = \frac{5500}{1 + \left(\frac{5500}{100} - 1\right)e^{-0.4t}} $$ Step 8: Graph the solution

Graph the solution

To graph the solution, create a plot of \(P(t)\) against \(t\). You should see a sigmoid curve, which represents the logistic growth model. The curve starts from the initial population \(P_{0}\), grows quickly as the population increases, and eventually reaches the carrying capacity \(K\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used extensively in modeling real-world situations where change is occurring. In the context of population growth, differential equations help describe how populations change over time.

One common form of differential equation for modeling population growth is the logistic growth equation. This equation takes into account factors like limited resources and competition, which naturally lead to the leveling off of population growth at a certain point. The logistic equation is given by:

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

Here, \(dP/dt\) represents the rate of change of the population \(P\) over time \(t\). The growth rate \(r\) influences how quickly the population grows, and \(K\) is the carrying capacity, or the maximum population size that the environment can sustainably support.

By solving this differential equation, we can predict the population size at any given time, thereby understanding the dynamic behavior of populations.

Initial Value Problem

An initial value problem in the context of differential equations refers to finding a specific solution that satisfies both the differential equation and an initial condition. In simple terms, it considers the initial state of the system at some moment, typically when \(t=0\).

For the problem outlined, the initial value is the population at time zero, denoted by \(P_{0}\). In our example, this is given as 100. This condition helps uniquely determine the solution to the differential equation. Without the initial value, you'd have a family of potential solutions, but the initial value pinpoints the exact one among them.

To solve an initial value problem, we:

• Write down the differential equation to reflect the situation.
• Apply integration to solve the equation, leveraging the initial value to find the constant of integration.
• Substitute known initial values to solve for unknown constants, completing the model.

Recognizing the initial state is crucial in understanding how a process will unfold over time.

Carrying Capacity

Carrying capacity is a key concept in ecological models, especially regarding population dynamics. It refers to the maximum number of individuals of a species that an environment can sustainably support without being degraded over time.

In the logistic growth model, carrying capacity \(K\) is a limiting factor. It keeps the population from growing indefinitely by establishing an upper bound. As the population approaches this limit, the growth rate slows and eventually stabilizes.

Carrying capacity can be influenced by various environmental factors such as:

• Availability of resources like food, water, and space.
• Predation pressures and disease.
• Competition among species.

By incorporating carrying capacity into models like the logistic growth model, scientists and researchers can more accurately simulate and predict how populations will change over time given constraints imposed by the environment. Understanding this concept is crucial not only in biology but also in managing and preserving ecosystems.