Problem 14 - Calculus

Question

Give the general solution to the logistic differential equation. $$\frac{d P}{d t}=0.012 P\left(1-\frac{P}{5700}\right)$$

Step-by-Step Solution

Answer

The general solution is \( P = \frac{5700De^{0.012t}}{1 + De^{0.012t}} \), where \( D \) is a constant.

Step 1: Identify the Logistic Differential Equation

The given differential equation is \( \frac{d P}{d t} = 0.012 P \left( 1 - \frac{P}{5700} \right) \). This is known as the logistic differential equation, which models population growth with a limiting factor, known as the carrying capacity.

Step 2: Separate Variables

The logistic differential equation can be rewritten in separable form as: \( \frac{1}{P(1 - \frac{P}{5700})} \frac{dP}{dt} = 0.012 \). Then, separate variables to get: \( \frac{dP}{P(1 - \frac{P}{5700})} = 0.012 \, dt \).

Step 3: Integrate Both Sides

Integrate the left-hand side with respect to \( P \) and the right-hand side with respect to \( t \). The left integral involves partial fraction decomposition: \( \int \frac{dP}{P} + \int \frac{1}{5700 - P} \, dP \). The right side integrates to \( 0.012t + C \), where \( C \) is the integration constant.

Step 4: Solve the Integral

The solution to the integrals are: \[ \ln|P| - \ln|5700 - P| = 0.012t + C \]Using logarithm properties, we can write:\[ \ln \left| \frac{P}{5700 - P} \right| = 0.012t + C \].

Step 5: Exponentiate to Solve for P

To solve for \( P \), exponentiate both sides: \[ \frac{P}{5700 - P} = e^{0.012t + C} = De^{0.012t} \]where \( D = e^C \) is an arbitrary constant. Rearrange this equation to solve for \( P \).

Step 6: Express P in Terms of t

Rearrange to express \( P \) explicitly in terms of \( t \): \[ P = \frac{5700De^{0.012t}}{1 + De^{0.012t}} \]This represents the general solution to the logistic equation in terms of \( P \) and \( t \).

Key Concepts

Population Growth Modeling

Population growth modeling is a mathematical way to predict how a population changes over time. For many species, including humans, population growth can often be best understood through differential equations. These equations account for ecological factors like food, space, and other resources that are available to the population.

In particular, the *logistic differential equation* is frequently used for this purpose. Unlike the simpler exponential growth model, which assumes unlimited growth, the logistic model accounts for constraints that naturally occur in the environment. This means that as the population increases, its growth rate decreases due to limitations in resources.

In the exercise we examined, the logistic differential equation is modeled by: \[\frac{d P}{d t} = 0.012 P \left(1-\frac{P}{5700}\right)\] This equation indicates that the growth rate - decreases as the population size \(P\) approaches the carrying capacity, - initially increases when the population is small,- and represents a more realistic scenario for how populations grow in the real world.

Such models are useful for ecologists and planners who want to make predictions about future population sizes.

Carrying Capacity

Carrying capacity is a fundamental concept closely tied to logistic population growth. It describes the maximum population size that an environment can sustain indefinitely. Essentially, it is the balance between the availability of resources and the needs of the inhabitants.

In the context of the logistic differential equation, the carrying capacity is represented by the number within the equation that stands as the threshold—here, it is 5700. The equation: \[\frac{d P}{d t} = 0.012 P \left(1-\frac{P}{5700}\right)\] shows that as \(P\) approaches 5700:- the growth rate slows down,- population growth levels off when population reaches the carrying capacity,- resources become limited, preventing further growth.

Understanding carrying capacity helps in realizing why we cannot expect a population to keep increasing endlessly. This has implications not just for ecological studies but also for sustainable development, helping societies manage resource allocation effectively.

Separation of Variables

Separation of variables is a mathematical method used to solve differential equations, such as the logistic differential equation. This technique involves rearranging the equation so that each side depends only on one variable.

For the logistic equation \(\frac{d P}{d t} = 0.012 P (1-\frac{P}{5700})\), the first step is rewriting it in separable form, isolating terms involving \(P\) and \(dt\):\[ \frac{1}{P(1 - \frac{P}{5700})} \frac{dP}{dt} = 0.012 \]By separating variables, you get:\[ \frac{dP}{P(1 - \frac{P}{5700})} = 0.012 \, dt \]

This allows each side to be integrated individually:- The left side with respect to \(P\),- The right side with respect to \(t\).

Integrating leads to: \[\ln \left| \frac{P}{5700 - P} \right| = 0.012t + C\]where \(C\) is the integration constant.

Using separation of variables, one can find solutions that describe how \(P\) changes over time, leading to more profound insights into the dynamics of the population under study.