Question

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

Short answer

\( y(t) = \frac{L}{1 + \left( \frac{L-y_0}{y_0} \right) e^{-kt}} \) is the solution.

Step-by-step solution

Understand the Logistic Differential Equation

The logistic differential equation is given as \( \frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right) \), where \(k\) is the constant of proportionality, \(L\) is the carrying capacity, and \(y\) is a function of time \(t\). Our goal is to solve this equation given the initial condition \(y(0) = y_0\).

Separate the Variables

Rearrange the equation to separate variables, allowing you to integrate both sides. Start by dividing both sides by \(y(1 - y/L)\) to get \( \frac{1}{y(1 - y/L)} \frac{dy}{dt} = k \). Then, separate the variables to get \( \frac{dy}{y(1 - y/L)} = k dt \).

Integrate Both Sides

Integrate both sides to solve for \(y\). The left side can be integrated by using partial fraction decomposition: \( \int \left( \frac{1}{y} + \frac{1}{L-y} \right) dy \) and the right side is simply \( \int k \ dt \). Solve the integrals: \( \ln|y| - \ln|L-y| = kt + C \).

Simplify the Solution

Rewrite the equation from the integration: \( \ln \left( \frac{y}{L-y} \right) = kt + C \). Exponentiate both sides to remove the natural logarithm: \( \frac{y}{L-y} = Ae^{kt} \), where \( A = e^C \).

Solve for the General Solution

Rearrange to solve for \(y\): \( y = \frac{L}{1 + (L/y_0 - 1)e^{-kt}} \), using the initial condition \( y(0) = y_0 \) to determine \( A \). Here, \( A = \frac{y_0}{L-y_0} \).

Verify the Solution

Substitute back the values to ensure the solution satisfies the original differential equation and the initial condition. Substitute \( t = 0 \) into the final expression for \(y\) and verify that \( y(0) = y_0 \).

Key concepts

Separation of Variables

Separation of variables is a technique used to solve differential equations. It's particularly useful for equations where variables can be rearranged and isolated on opposite sides of the equation. This approach allows each side to involve only one variable, making it easier to integrate.

For the logistic differential equation \( \frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right) \), we begin by separating the variables. We rearrange this into the form \( \frac{dy}{y(1 - y/L)} = k \, dt \).

* Steps to separate the variables:

• Multiply or divide both sides of the equation to isolate terms involving \(y\) on one side and \(t\) on the other.
• This often involves dividing by \(y\) or some function of \(y\).
• The goal is to make both sides of the equation integrable in terms of their respective variables.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify a complex rational expression into simpler, easier components. This is especially useful when integrating expressions obtained after using the separation of variables technique.

In our situation, for the equation \( \frac{dy}{y(1 - y/L)} = k \, dt \), we utilize partial fraction decomposition after separation. This involves expressing \( \frac{1}{y(1 - y/L)} \) as the sum of simpler fractions:

* \( \frac{1}{y} + \frac{1}{L-y} \)

This allows each component to be integrated separately. The advantage here is that simple logarithmic functions arise when integrating these types of fractional expressions, leading to the result \( \ln|y| - \ln|L-y| \), after integration.

* Points to remember:

• Differentiate each fraction to ensure they sum back to the original expression.
• Verify your decomposition before integrating, as incorrect fractions will yield inaccurate results.

Carrying Capacity

Carrying capacity is a crucial concept when dealing with logistic models. It represents the maximum population size that an environment can sustain indefinitely, given the resources, space, and other ecological factors available.

In the logistic differential equation \( \frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right) \), the parameter \( L \) denotes the carrying capacity.

* Why it matters:

• It limits exponential growth as \( y \) approaches \( L \).
• Induces a natural "stabilization" effect in the model, preventing uncontrolled growth.

As the population \( y \) nears \( L \), growth slows, which balances the environment's capacity. This logistic growth pattern reflects many real-world population dynamics more accurately than simple exponential models.

Integration of Differential Equations

Integration is the crucial step in solving differential equations once variables have been separated. Through integration, we find the function that describes the change happening in the system.

After separating and expressing the logistic differential equation \( \frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right) \) into integrable forms, we move to:* Solve the left side with care, especially using the results from partial fraction decomposition.* Typically, expressions of the form \( \ln|y| - \ln|L-y| \) emerge.* Directly integrate the right side: \( kt \), representing the constant growth over time.

As a result, our integrated forms yield the relationship:\[ \ln \left( \frac{y}{L-y} \right) = kt + C \]

From this point, the process involves applying exponential functions to solve explicitly for \( y \), forming the complete solution to our logistic equation with specified initial conditions. Integration thus transforms complex equations into applicable solutions reflecting real-world phenomena. It provides essential insights into how different variables interact over time, especially in dynamic systems like populations or resources.