/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 In Exercises \(1-8\) a. Identi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 7

In Exercises \(1-8\) a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$ y^{\prime}=(y-1)(y-2)(y-3) $$

Short Answer

Expert verified
Equilibrium values: \( y=1, 2, 3 \). \( y=2 \) is stable.

Step by step solution

01

Identify Equilibrium Values

Equilibrium values occur where \( y' = 0 \). Set the equation \( (y-1)(y-2)(y-3) = 0 \). The solutions are \( y = 1, 2, 3 \). These are the equilibrium values.
02

Determine Stability of Equilibrium Values

Analyze the sign of \( y' = (y-1)(y-2)(y-3) \) around each equilibrium point. - For \( y < 1 \), all factors are negative making \( y' > 0 \). - Just above \( y=1 \), plug \( y=1.5 \), we have \( y' > 0 \). - Between \( y=2 \) and \( y=3 \), plug \( y=2.5 \), we have \( y' < 0 \).- For \( y > 3 \), all factors are positive making \( y' > 0 \). Therefore, \( y=1 \) is unstable, \( y=2 \) is stable, and \( y=3 \) is unstable.
03

Construct a Phase Line

Draw a vertical line representing the values of \( y \). Mark the equilibrium points \( y=1, 2, \) and \( 3 \). Indicate the sign of \( y' \) by arrows: - For \( y < 1 \) and \( y > 3 \), point arrows upward (\( y' > 0 \)). - Between \( y=1 \) and \( y=2 \), and \( y>3\), arrows point downward (\( y' < 0 \)). - Between \( y=2 \) and \( y=3 \), point arrows upward (\( y' > 0 \)).
04

Analyze Second Derivative for Concavity

Find the derivative of \( y' \, \Rightarrow \, y'' = 3y^2 - 12y + 11 \). Analyze concavity: - Between each pair of equilibria, substitute values to determine sign of \( y'' \).- Example between \( y=1 \) and \( y=2 \): plug \( y=1.5 \) yields \( y'' < 0 \). Therefore check each region around points \( y=1, 2, \) and \( 3 \) to confirm.
05

Sketch Solution Curves

Use the phase line and sign of \( y'' \) to draw solution curves.- Solutions starting below \( y=1 \) or above \( y=3 \) tend away from equilibrium points.- The curve flattens near \( y=2 \) indicating stability, and solutions tend towards it. Sketch various solutions, noting they converge at equilibrium for \( y=2 \) and diverge at other points.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equilibrium Values
In differential equations, equilibrium values are crucial as they represent the points where the solution doesn't change. These are the points where the rate of change, denoted as \( y' \), equals zero. For the given exercise, you need to solve the equation \( (y-1)(y-2)(y-3) = 0 \) to find these values. Here, the solutions are \( y = 1, 2, 3 \). Each of these is an equilibrium point because substituting them into \( y' \) results in zero.
This means when the system reaches these points, there's no tendency to move in either direction at that instant.
Equilibrium values are the foundation for exploring the system's behavior and are essential in determining stability.
Phase Line
A phase line is a simple tool that helps visualize the behavior of a differential equation without needing a full graph. It's essentially a vertical line where you mark the equilibrium points that we found in the previous step at \( y = 1, 2, 3 \).
Next, add arrows to this line to indicate where the derivative \( y' \) is positive or negative. For our exercise:
  • For \( y < 1 \) and \( y > 3 \), the phase line shows upward arrows (\( y' > 0 \)).
  • Between \( y = 1 \) and \( y = 2 \), the arrows point downward (\( y' < 0 \)).
  • Finally, between \( y = 2 \) and \( y = 3 \), the arrows point upward (\( y' > 0 \)).

Phase lines help quickly assess the stability and predict the trajectories of solutions.
Stability Analysis
Stability analysis focuses on understanding whether solutions to a differential equation tend to return to equilibrium points after small perturbations. Each equilibrium value can be stable, unstable, or semi-stable.
  • A stable equilibrium point means small deviations diminish over time, driving solutions back toward the point.
  • An unstable equilibrium means deviations grow, and solutions move away from the point.
For the exercise, an analysis of the sign of \( y' \) around each equilibrium value shows:
  • \( y = 1 \): Since \( y' \) is positive on both sides, it's unstable; solutions diverge.
  • \( y = 2 \): With \( y' \) negative before and positive after, this makes \( y = 2 \) stable; it's a point of attraction.
  • \( y = 3 \): As \( y' \) is positive on the right, it is unstable as well.

This analysis is vital for predicting the long-term behavior of solutions.
Solution Curves
Solution curves provide a visual representation of how solutions change over time concerning the differential equation. By using phase lines as your guide and knowing the concavity from \( y'' \), you can sketch the behavior of solutions.
  • Solutions beginning below \( y = 1 \) or above \( y = 3 \) will diverge away from these points due to instability.
  • Solutions that start near \( y = 2 \) will converge towards it, depicting its stability as solutions flatten out.
Sketching these curves illustrates how the system evolves, dynamically stabilizing at \( y = 2 \) or drifting away from \( y = 1 \) and \( y = 3 \).
This method connects abstract mathematical concepts with tangible visuals, aiding comprehensive understanding.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The spread of information Sociologists recognize a phenome- non called social diffusion, which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it. If \(X\) denotes the number of individuals who have the information in a population of \(N\) people, then a mathematical model for social diffusion is given by $$ \frac{d X}{d t}=k X(N-X) $$ where \(t\) represents time in days and \(k\) is a positive constant. a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of \(X^{\prime}\) and \(X^{\prime \prime}\) . c. Sketch representative solution curves. d. Predict the value of \(X\) for which the information is spreading most rapidly. How many people eventually receive the information?

Solve the differential equations. \(x \frac{d y}{d x}=\frac{\cos x}{x}-2 y, \quad x>0\)

In Exercises \(13-16,\) use Euler's method with the specified step size to estimate the value of the solution at the given point \(x^{*} .\) Find the value of the exact solution at \(x^{*}\) . $$ y^{\prime}=\sqrt{x} / y, \quad y>0, \quad y(0)=1, \quad d x=0.1, \quad x^{*}=1 $$

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y-\) window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b] .\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {0 \leq x \leq 4, \quad 0 \leq y \leq 3 ; \quad b=3}\end{array} $$

In Exercises \(1-6,\) use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$ y^{\prime}=y+e^{x}-2, \quad y(0)=2, \quad d x=0.5 $$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.