Question

Solve the logistic differential equation representing population growth with the given initial condition. Then use the solution to predict the population size at time \(t=3 .\) \(y^{\prime}=\frac{1}{10} y(12-y), y(0)=2\)

Short answer

The population size at \( t = 3 \) is approximately 7.78.

Step-by-step solution

Identify the Logistic Equation

The given logistic differential equation is \( y' = \frac{1}{10} y(12 - y) \). This is a typical logistic equation with growth rate \( r = \frac{1}{10} \) and carrying capacity \( K = 12 \).

Write the General Solution

The general solution for the logistic differential equation \( \frac{dy}{dt} = ry(K-y) \) is given by: \[ y(t) = \frac{K}{1 + Ce^{-rKt}} \] where \( C \) is a constant determined by the initial condition.

Apply the Initial Condition

Given the initial condition \( y(0) = 2 \), substitute \( t = 0 \) and \( y = 2 \) into the general solution: \[ 2 = \frac{12}{1 + C} \]. Solve for \( C \), yielding \( C = 5 \).

Write the Particular Solution

Substitute \( C = 5 \) back into the general solution to get the particular solution: \[ y(t) = \frac{12}{1 + 5e^{-\frac{12}{10}t}} \].

Predict the Population at \( t = 3 \)

Substitute \( t = 3 \) into the particular solution: \[ y(3) = \frac{12}{1 + 5e^{-\frac{36}{10}}} \]. Calculate this expression to get the predicted value of \( y(3) \).

Calculate and Interpret

Compute \( y(3) \) using the expression obtained in Step 5: \( y(3) = \frac{12}{1 + 5e^{-3.6}} \). After calculating, we find \( y(3) \approx 7.78 \). This represents the predicted population size at \( t = 3 \).

Key concepts

population growth

Population growth, in the context of differential equations, describes how the number of individuals in a population changes over time. In many situations, population growth does not occur indefinitely at an increasing rate. Instead, it experiences initial rapid growth, which slows over time due to various limiting factors like resources and space. This gradual slowing is effectively captured by the logistic model.

The logistic differential equation, such as the one in this problem, models realistic population dynamics. It includes terms for both the natural exponential growth of the population and the limiting factors due to limited resources. In our exercise, this is represented by the equation \( y' = \frac{1}{10} y(12-y) \), where \( y' \) is the rate of change of population size with respect to time. Here, 1/10 is the growth rate, and 12 is the maximum population that the environment can sustain, known as the carrying capacity.

initial condition

The initial condition of a differential equation determines the starting point for solving the equation. It's essential because it allows us to find a specific solution that fits the initial constraints of the model. In this case, the initial condition given is \( y(0) = 2 \), meaning the population size at time zero is 2.

Initial conditions help us calculate a constant in the logistic equation solution. By substituting the initial condition values into the general solution, we can solve for the constant and thus obtain a particular solution that describes the population's behavior over time under the given initial scenario.

carrying capacity

Carrying capacity is a concept integral to understanding population dynamics. It's defined as the maximum population size that an environment can sustain indefinitely. For the logistic model, it's the plateau level the population approaches as time progresses, due to limitations like food, mates, and space.

In our example, the carrying capacity is 12. This means that even if there are favorable conditions initially, the population will not exceed this size permanently. The logistic model uses the carrying capacity to adjust the growth rate, ultimately balancing the birth and death rates as resources become limited.

logistic model solutions

Logistic model solutions provide us with a way to describe how populations evolve over time, considering both the growth dynamics and the resource constraints. We start with the general solution of the logistic differential equation:

\[ y(t) = \frac{K}{1 + Ce^{-rKt}} \]

This captures both exponential growth and leveling off due to the carrying capacity \( K \). The constant \( C \) is determined using the initial condition, ensuring the solution starts at the right population size.

With the initial condition provided as \( y(0) = 2 \), we solve for \( C \) to find a particular solution:

\[ y(t) = \frac{12}{1 + 5e^{-\frac{12}{10}t}} \]

This particular solution tells us exactly how the population will evolve, accounting for both the inherent growth rate and the environmental limitations, giving us meaningful predictions and insights into future population sizes, such as at \( t = 3 \). By substituting \( t = 3 \), we find that the predicted population is about 7.78, showing how the logistic model effectively captures the growth pattern.