Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

The equation for logistic growth is $$ \frac{d y}{d t}=k y(L-y) $$ Show that this differential equation has the solution $$ y=\frac{L y_{0}}{y_{0}+\left(L-y_{0}\right) e^{-L k t}} $$

Step-by-Step Solution

Answer

The solution is verified by checking both the differential equation and the initial condition.

Step 1: Identification of Variables

Identify that in the logistic growth equation \( \frac{d y}{d t} = k y (L-y) \), \( y \) is the function of \( t \), \( L \) is the carrying capacity, \( k \) is the growth rate constant, and \( y_0 \) is the initial population size.

Step 2: Assume a Solution Form

Assume a solution in the form \( y(t) = \frac{L y_{0}}{y_{0}+eta} \) where \( \beta = C \cdot e^{-L k t} \) and \( C \) is a constant to be determined.

Step 3: Differentiate Assumed Solution

Differentiate the assumed form \( y(t) = \frac{L y_{0}}{y_{0} + (L - y_0)e^{-L k t}} \) with respect to \( t \) to find \( \frac{dy}{dt} \). Apply the chain rule carefully to find the expression for \( \frac{dy}{dt} \).

Step 4: Substitute Expression into Logistic Equation

Substitute the differentiated expression \( \frac{dy}{dt} \) back into the original logistic equation \( \frac{d y}{d t}=k y(L-y) \) and simplify. Ensure each side of the equation leads to the equality to verify it satisfies the differential equation.

Step 5: Verify Initial Condition

Check if \( y(t) \) satisfies the initial condition at \( t = 0 \). Substitute \( t = 0 \) into \( y(t) = \frac{L y_{0}}{y_{0}+(L-y_{0})e^{-Lkt}} \) which simplifies to \( y(0) = y_0 \). This confirms that the assumed solution satisfies the initial condition.

Step 6: Conclusion

Conclude that because the assumed solution satisfies the differential equation and the initial condition, \( y(t) = \frac{L y_{0}}{y_{0}+(L-y_{0})e^{-Lkt}} \) is indeed the correct solution to the logistic differential equation.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. In our context, the logistic growth differential equation is given by \[ \frac{d y}{d t} = k y (L-y) \] This equation describes how a quantity, like a population, changes over time based on its current value. It includes the growth rate constant, \( k \), and considers the carrying capacity, \( L \).

The purpose of such an equation is to model the rate at which a population grows in an environment that has a natural limit. It communicates how the growth speed depends on how far the current population is from its maximum capacity.

• If the population \( y \) is much lower than the carrying capacity \( L \), the growth rate is almost exponential.
• As \( y \) approaches \( L \), the growth rate slows down significantly.

Solving the logistic differential equation provides a function \( y(t) \) that predicts population size at any time \( t \). This is critical for understanding and predicting population dynamics.

Carrying Capacity

Carrying capacity, denoted \( L \) in the logistic growth model, represents the maximum number of individuals that an environment can sustainably support. Imagine a lake with enough resources to support only 100 fish.

In the equation \[ \frac{d y}{d t} = k y (L-y), \] carrying capacity plays a dual role.

• It limits growth as the population size \( y \) approaches \( L \), stabilizing the population.
• It forms the threshold that regulates growth speed - the closer \( y \) gets to \( L \), the slower the growth becomes.

Understanding \( L \) in real-world applications means grasping how space, food, and other resources constrain population expansion. This parameter guides conservation efforts and sustainable resource management.

Growth Rate Constant

The growth rate constant, denoted \( k \), is a crucial part of the logistic growth equation. It signifies how fast the population grows over time. In the model \[ \frac{d y}{d t} = k y (L-y), \] \( k \) essentially scales the growth rate and influences the responsiveness of the population to changes in size and resources.

A larger \( k \) means rapid adaptation to favorable conditions, leading to swift population changes. Conversely, a smaller \( k \) indicates a slow response, causing gradual growth.

• With a larger \( k \), populations reach carrying capacity more quickly.
• A smaller \( k \) leads to slower adjustment periods, affecting long-term dynamics.

This parameter depends on biological characteristics of the species and environmental factors, making it vital in ecological modeling and population management.

Initial Population Size

The initial population size, represented as \( y_0 \), is the population's starting point at time \( t = 0 \). In the logistic growth equation and its solution \[ y(t) = \frac{L y_{0}}{y_{0}+(L-y_{0})e^{-Lkt}}, \] \( y_0 \) influences how quickly a population approaches its carrying capacity \( L \) based on initial conditions.

By knowing \( y_0 \), one can determine how a new environment or an existing population will grow over time, given no disruptive circumstances occur.

• It provides the basis for all predictive modeling. If \( y_0 \) is small compared to \( L \), the population grows rapidly at first.
• Larger \( y_0 \) values close to \( L \) mean slower initial growth since the population is already near the maximum sustainable size.

Understanding this parameter helps in deploying new species into habitats and planning sustainable practices in resource utilization.