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Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. $$ \begin{array}{l} y^{\prime}=y(2-y) \\ y(0)=4 \end{array} $$

Short Answer

Expert verified
The solution is \( y(t) = \frac{4}{2 - e^{-2t}} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is \( y' = y(2 - y) \). This resembles the general form of a logistic growth model, which is \( y' = ry(1 - \frac{y}{K}) \), where \( r \) is the rate of growth and \( K \) is the carrying capacity.
02

Determine Growth Rate and Carrying Capacity

Compare the equation \( y' = y(2 - y) \) with the logistic growth equation \( y' = ry(1 - \frac{y}{K}) \). Here, \( r = 2 \) and \( K = 2 \). This means the growth rate is 2 and the carrying capacity is 2.
03

Use the Logistic Growth Solution Formula

The solution to the logistic differential equation is \( y(t) = \frac{K}{1+Ce^{-rt}} \). We need to determine the constant \( C \) using the initial condition.
04

Substitute Initial Condition to Find \( C \)

Use the initial condition \( y(0) = 4 \). Substitute into the logistic growth formula: \( 4 = \frac{2}{1+Ce^{0}} \). Simplifying gives \( 4 = \frac{2}{1+C} \). Solve for \( C \) by rearranging: \( 1 + C = \frac{1}{2} \), leading to \( C = -\frac{1}{2} \).
05

Write the Full Solution

Substitute \( C = -\frac{1}{2} \) back into the equation \( y(t) = \frac{2}{1+Ce^{-2t}} \) to get \( y(t) = \frac{2}{1-\frac{1}{2}e^{-2t}} \). Simplify the fraction: \[ y(t) = \frac{2}{1 - \frac{1}{2} e^{-2t}} = \frac{2}{\frac{1\cdot2 - e^{-2t}}{2}} = \frac{4}{2 - e^{-2t}} \]. The solution is \( y(t) = \frac{4}{2 - e^{-2t}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are fundamental in modeling how systems change over time. They express relationships involving rates of change. In this exercise, the differential equation is \( y' = y(2-y) \). This is an example of a first-order equation where \( y' \) represents the derivative of \( y \) with respect to time \( t \).

This equation is recognized as a form of the logistic growth model, a non-linear differential equation. By comparing it to the standard form of a logistic equation \( y' = ry(1 - \frac{y}{K}) \), we can see how the terms relate to growth factors. The coefficient before \( y(2-y) \) provides insights into natural limits on the system's growth, manipulating \( y \) so that it trends towards a finite limit (the carrying capacity).

Solving a logistic differential equation often involves finding explicit solutions, which include constants determined by initial conditions. Here, the initial condition is \( y(0) = 4 \), and such conditions help determine specific constants like \( C \) in the function \( y(t) \). Understanding differential equations allows us to predict how populations or other systems evolve over time.
Growth Rate
Growth rate in the context of differential equations reflects how fast the quantity changes over time. In the logistic growth model, it is identified by \( r \), known as the intrinsic growth rate. This value represents the potential rate of increase per individual in an ideal, unlimited environment.

In the given differential equation, \( y' = y(2-y) \), the growth rate \( r \) is directly inferred. From comparison with the logistic model form, \( r \) is found to be 2. This indicates the maximum proportional increase in the population when it is very small, suggesting rapid initial growth.

However, it is crucial to note that this rate is modulated by the factor \( (2-y) \), which incorporates the effects of competition or crowding as the population \( y \) approaches its carrying capacity. As \( y \) grows, the term \( (2-y) \) acts to slow the growth, reflecting the decreasing resources or space as the population nears its environment's sustainable limit.
Carrying Capacity
The concept of carrying capacity \( K \) is integral to understanding logistic growth models. It refers to the maximum population size that an environment can sustain indefinitely. Beyond this limit, populations cannot increase sustainably due to resource limitations, leading to stabilization or decline.

In the logistic differential equation \( y' = y(2-y) \), the carrying capacity is identified as \( K = 2 \) by comparing it with the standard logistic model \( y' = ry(1 - \frac{y}{K}) \). This value signifies the threshold that the population approaches as time progresses, influencing the shape of the logistic curve.

As a biological or ecological constraint, carrying capacity ensures that the population does not exceed what the environment can support. As the population approaches \( K \), growth slows down significantly, resulting in a stable population level. Recognizing and computing \( K \) in logistic growth problems help predict long-term population or resource dynamics.

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Most popular questions from this chapter

Think of the slope field for the differential equation \(\frac{d y}{d x}=\frac{6 x}{y^{2}} .\) What is the sign of the slope in quadrant \(I\) (where \(x\) and \(y\) are both positive)? What is the sign of the slope in each of the other three quadrants? Check your answers by looking at the slope field on page 594.

A Ponzi scheme is an investment fraud that promises high returns but in which funds, instead of being invested, are merely used to pay returns to the investors and profits to the fund manager. To avoid running out of money, new investors must be brought in at an increasing rate to provide funds to pay off existing investors. Eventually the scheme must collapse in debt when not enough new investors can be found. Suppose that each of 10 investors deposits \(\$ 100,000\) into a fund and is promised a \(20 \%\) annual return. However, the \(\$ 1,000,000\) collected is used to pay each investor the required \(\$ 20,000\) return, with the remaining \(\$ 800,000\) kept by the fund manager. Let \(y(t)\) be the total number of investors needed after \(t\) years so that incoming funds will be enough to pay the existing investors plus the manager's \(\$ 800,000 .\) Representing dollar amounts in thousands, we have $$ \begin{array}{l} \left(\begin{array}{c} \text { Annual } \\ \text { inflow } \end{array}\right)=100 \cdot \frac{d y}{d t} \\ \left(\begin{array}{l} \text { Annual } \\ \text { outflow } \end{array}\right)=20 \cdot y+800 \end{array} $$ For inflow to equal outflow, we must have the differential equation and initial condition $$ \left\\{\begin{array}{l} 100 \cdot \frac{d y}{d t}=20 y+800 \\ y(0)=10 \end{array}\right. $$ a. Solve this differential equation and initial condition. b. Use your solution to find how many investors would be needed after 10 years, after 20 years, after 30 years, and after 50 years.

Solve each first-order linear differential equation. $$ y^{\prime}=x+y $$

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=a x^{2} y \\ y(0)=2 \end{array} \quad(\text { for constant } a>0)\right.$$

For each initial value problem, calculate the Euler approximation for the solution on the interval [0,1] using \(n=4\) segments. Draw the graph of your approximation. (Carry out the calculations "by hand" with the aid of a calculator, rounding to two decimal places. Answers may differ slightly, depending on when you do the rounding.) $$ \begin{array}{l} y^{\prime}+2 y=e^{4 x} \\ y(0)=2 \end{array} $$

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