Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the logistic differential equation for an arbitrary constant of proportionality \(k\), capacity \(L\), and initial condition \(y(0)=y_{0}\).

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{L}{1 + (\frac{L-y_{0}}{y_{0}})e^{-kt}} \).

Step 1: Write the Logistic Differential Equation

The logistic differential equation is expressed as \( \frac{dy}{dt} = k y (L-y) \), where \( k \) is the constant proportionality, \( L \) is the carrying capacity, and \( y_0 \) is the initial condition \( y(0) = y_0 \).

Step 2: Separate Variables

To solve, separate variables: \( \frac{dy}{y(L-y)} = k dt \). This allows us to integrate both sides separately with respect to their variable.

Step 3: Integrate Both Sides

To integrate the left side, use partial fraction decomposition: \( \frac{1}{y(L-y)} = \frac{A}{y} + \frac{B}{L-y} \). Solving for \( A \) and \( B \), we have: \( \frac{1}{y} - \frac{1}{L-y} \). Integrate: \( \int \left(\frac{1}{y} - \frac{1}{L-y}\right) dy = \int k \, dt \).

Step 4: Determine the Integrals

The integral of the left side is \( \ln|y| - \ln|L-y| \). The integral of the right side is \( kt + C \), where \( C \) is the constant of integration.

Step 5: Solve for \( y \)

Combine results by exponentiating both sides to remove logarithms: \( \frac{y}{L-y} = Ce^{kt} \). Solve for \( y \): \( y = \frac{L}{1 + De^{-kt}} \), where \( D = \frac{L-y_{0}}{y_{0}} \).

Step 6: Apply the Initial Condition

Use the initial condition \( y(0) = y_0 \) to determine \( D \): \( y = \frac{L}{1 + \left(\frac{L-y_{0}}{y_{0}}\right)e^{-kt}} \).

Step 7: Conclusion

The solution of the logistic differential equation is \( y(t) = \frac{L}{1 + \left(\frac{L-y_{0}}{y_{0}}\right)e^{-kt}} \), where \( L \) is the carrying capacity, \( k \) is the rate constant, and \( y_0 \) is the initial population.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify the integration process, especially helpful for complex rational expressions. In the context of logistic differential equations, it allows us to break down a fraction into simpler parts that are easier to integrate.
For instance, when working with the term \( \frac{1}{y(L-y)} \), we can express it as a sum of partial fractions: \( \frac{A}{y} + \frac{B}{L-y} \). By finding values for \( A \) and \( B \), we simplify the integration process.

• First, set up the equation: \( \frac{1}{y(L-y)} = \frac{A}{y} + \frac{B}{L-y} \)
• Then solve for \( A \) and \( B \), which typically involves algebraic manipulation and setting up equations.
• The result helps us rewrite the integral as: \( \int \left( \frac{1}{y} - \frac{1}{L-y} \right) dy \).

This simplification is key because integrating \( \frac{1}{y} \) and \( \frac{1}{L-y} \) is more straightforward. It transforms the problem into manageable parts, making the entire process of solving differential equations less cumbersome.

Variable Separation

Variable separation is a technique used in solving differential equations where all terms involving one variable are moved to one side of the equation, and terms involving the other variable are kept on the opposite side.
In the logistic differential equation \( \frac{dy}{dt} = k y (L-y) \), our goal is to separate \( y \) and \( t \). By dividing both sides by \( y(L-y) \), and multiplying both sides by \( dt \), we isolate the variables:

• Left side: \( \frac{dy}{y(L-y)} \)
• Right side: \( k \, dt \)

Once separated, both sides can be integrated independently. This method is crucial as it simplifies the process of solving otherwise complex equations. Variable separation thus aids in systematically tackling the solution by handling each part of the equation individually.

Carrying Capacity

Carrying capacity, denoted as \( L \), is a central concept in the logistic model. It reflects the maximum population size that an environment can support indefinitely without ecological damage.
In the logistic differential equation, carrying capacity determines the limit of growth. As \( y \), the population, approaches \( L \), the growth rate decreases, modeling real-world scenarios like limited resources or environmental constraints.

• When \( y \) is much smaller than \( L \), the population grows exponentially.
• As \( y \) nears \( L \), the growth rate slows due to diminishing available resources.
• Finally, once \( y \) reaches \( L \), growth stops, and the population stabilizes.

Understanding carrying capacity helps predict population dynamics within ecosystems, making it an essential biological concept captured by the logistic model.

Initial Condition

An initial condition is a specific value that defines the state of a system at the start of observation. In differential equations, it provides necessary information to find particular solutions that match given scenarios.
For the logistic equation, the initial condition typically gives the initial population size \( y_0 \) at time \( t = 0 \). This helps to determine unknown constants in the solution.

• Applying the initial condition \( y(0) = y_0 \) allows the calculation of constants like \( D \) in the solution formula.
• It personalizes the general solution to reflect real-world values and scenarios.
• Ensures that the solution curve starts at the correct initial point on the population growth graph.

By using initial conditions, we make the model applicable to practical problems, effectively adapting it to specific cases like population studies or resource consumption models.