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Chapter 9: Problem 21

Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. $$ \begin{array}{l} y^{\prime}=0.05(0.25-y) \\ y(0)=0 \end{array} $$

Short Answer

Expert verified
The solution is \( y(t) = 0.25 - 0.25e^{-0.05t} \).

Step by step solution

01

Identify the type of growth

The differential equation \( y' = 0.05(0.25 - y) \) is a linear differential equation that describes a limited growth model. The term \( 0.25 - y \) suggests that the growth rate decreases as \( y \) approaches 0.25, indicating a carrying capacity.
02

Write the standard form of limited growth equation

The standard form for a differential equation describing limited growth is \( y' = r(K-y) \), where \( K \) is the carrying capacity and \( r \) is the growth rate. Comparing this with the given equation, we have \( r = 0.05 \) and \( K = 0.25 \).
03

Solve the differential equation

To solve the equation, we separate the variables: \( \frac{dy}{0.25-y} = 0.05 \, dt \). Integrating both sides gives \( -\ln|0.25-y| = 0.05t + C \). By exponentiating both sides, we find \( 0.25 - y = Ae^{-0.05t} \), where \( A = e^C \) is the integration constant.
04

Use the initial condition to find integration constant

Use the initial condition \( y(0) = 0 \) to determine \( A \). Substitute \( t = 0 \) and \( y = 0 \) into the equation: \( 0.25 - 0 = A \cdot e^{0} \). Thus, \( A = 0.25 \).
05

Write the solution function

Substitute \( A = 0.25 \) back into the solution equation: \( 0.25 - y = 0.25e^{-0.05t} \). Solving for \( y \), we get \( y = 0.25 - 0.25e^{-0.05t} \).

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Key Concepts

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

Differential Equation
A differential equation is an equation that relates a function with its derivatives. In the context of growth models, differential equations are used to describe how a quantity such as population, height, or mass changes over time. In our exercise, the differential equation is given as \( y' = 0.05(0.25 - y) \). This equation shows how the variable \( y \) (which can represent a population or another quantity) changes with respect to time \( t \). Breaking it down:
  • The left side, \( y' \), represents the rate of change of \( y \) with respect to time.
  • The right side, \( 0.05(0.25 - y) \), gives the specific form of the rate of change, depending on \( y \).
In this equation, the impact of \( y \) on its growth rate is negative, meaning as \( y \) increases, the rate of its growth decreases. This is characteristic of limited growth.
Carrying Capacity
Carrying capacity is a concept that refers to the maximum population size that an environment can sustain indefinitely. In a mathematical model describing population growth, the carrying capacity is a limit that the population approaches over time. In our exercise, the carrying capacity \( K \) is indicated by the constant \( 0.25 \) in the differential equation \( y' = 0.05(0.25 - y) \). Here, the term \( 0.25 \) suggests that the variable \( y \) will grow until it reaches a value of \( 0.25 \). Key points about carrying capacity:
  • It is the point where the rate of growth becomes zero.
  • The quantity \( y \) does not exceed this value under the model's assumptions.
  • It ensures that resources or limitations are accounted for in the population growth model.
By reflecting environmental limitations, the carrying capacity is crucial in limiting growth to realistic levels.
Growth Rate
Growth rate in a differential equation describes how fast a quantity is increasing or decreasing over time. In the given equation \( y' = 0.05(0.25 - y) \), the growth rate is represented by the constant \( 0.05 \). This constant tells us how quickly the variable \( y \) approaches its carrying capacity.Considerations of the growth rate:
  • A larger growth rate implies a steeper, faster increase.
  • A smaller rate means a slower approach to the carrying capacity.
  • The growth rate in this model is constant, meaning it does not change as \( y \) changes.
A critical part of analyzing such equations is understanding how different growth rates affect the time taken to reach or approach the carrying capacity.
Initial Condition
The initial condition of a differential equation specifies the value of the function at a particular point, often used to find the constant of integration when solving the equation. In our context, the initial condition is \( y(0) = 0 \), meaning that at time \( t = 0 \), the quantity \( y \) is equal to \( 0 \).Why initial conditions matter:
  • They provide a starting point for the model, giving context to the solution.
  • They are essential for determining specific solutions from a family of possible solutions.
  • With the initial condition \( y(0) = 0 \), we determine the integration constant \( A \).
In our problem, applying this initial condition results in the integration constant being \( A = 0.25 \), completing the solution to the differential equation.

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Most popular questions from this chapter

Let \(y(t)\) be the value of a commercial building (in millions of dollars) after \(t\) years. a. Write a differential equation that says that the rate of growth of the value of the building is equal to two times the one-half power of its present value. b. Write an initial condition that says that at time zero the value of the building is 9 million dollars. c. Solve the differential equation and initial condition. d. Use your solution to find the value of the building at time \(t=5\) years.

For each initial value problem: a. Use an Euler's method graphing calculator program to find the estimate for \(y(2)\). Use the interval [0,2] with \(n=50\) segments. b. Solve the differential equation and initial condition exactly by separating variables or using an integrating factor. c. Evaluate the solution that you found in part (b) at \(x=2\). Compare this actual value of \(y(2)\) with the estimate of \(y(2)\) that you found in part (a). $$ \begin{array}{l} y^{\prime}+y=2 e^{x} \\ y(0)=5 \end{array} $$

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(y^{\prime}=e^{x / y}\) \(y(1)=0.4\) Approximate the solution at \(x=3\)

Solve each by the appropriate technique. a. \(y y^{\prime}=x\) b. \(y^{\prime}+y=e^{-x}\)

The following are differential equations stated in words. Find the general solution of each. The derivative of a function at each point is 6 .
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