/*! 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 4 Find all equilibria of the auton... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 4

Find all equilibria of the autonomous differential equation and construct the phase plot. (a) $$ y^{\prime}=y^{2}-2$$ (b)$$y^{\prime}=\frac{y-3}{y+9}, \quad y \geqslant 0$$ (c) $$y^{\prime}=y(3-y)\left(25-y^{2}\right)$$

Short Answer

Expert verified
Equilibria: (a) \( y=\pm\sqrt{2} \), (b) \( y=3 \), (c) \( y=0,3,5,-5 \).

Step by step solution

01

Understanding Equilibrium

For any differential equation \( y' = f(y) \), equilibria occur where \( y' = 0 \), meaning \( f(y) = 0 \). We will determine the values of \( y \) that satisfy this condition for each given equation.
02

Solving for Equilibria (a)

For the equation \( y' = y^2 - 2 \), set this equal to zero: \( y^2 - 2 = 0 \). Solve for \( y \) to find \( y^2 = 2 \). Thus, \( y = \pm \sqrt{2} \) are the equilibria.
03

Solving for Equilibria (b)

For the equation \( y' = \frac{y-3}{y+9} \), set the numerator equal to zero: \( y - 3 = 0 \). Solve for \( y \) to find \( y = 3 \). This is the equilibrium point; note \( y \geq 0 \) does not restrict this solution.
04

Solving for Equilibria (c)

For the equation \( y' = y(3-y)(25-y^2) \), set each factor to zero: \( y = 0 \), \( 3 - y = 0 \) giving \( y = 3 \), and \( 25 - y^2 = 0 \) giving \( y = \pm 5 \). The equilibria are \( y = 0, 3, 5, -5 \).
05

Constructing the Phase Plot (a)

Consider the intervals determined by the roots \( y = \pm \sqrt{2} \). For \( y' = y^2 - 2 \), check test points in \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, \infty)\). The derivative is positive in \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\), negative in \((-\sqrt{2}, \sqrt{2})\). This yields motion direction for the phase plot.
06

Constructing the Phase Plot (b)

The function is defined and smoothly differentiable for all \( y \geq 0 \) except \( y = -9 \), which is not considered in the non-negative domain. For \( y' = \frac{y-3}{y+9} \), test points around \( y = 3 \). The derivative is positive for \( y > 3 \), negative for \( y < 3 \), leading to a stable equilibrium at \( y = 3 \).
07

Constructing the Phase Plot (c)

Consider intervals \((-\infty, -5)\), \((-5, 0)\), \((0, 3)\), \((3, 5)\), and \((5, \infty)\). Evaluate signs of \( y' \) in each. \( y' \) is negative in \((-\infty, -5)\), positive in \((-5, 0)\) and \((0, 3)\), negative in \((3, 5)\), and positive in \((5, \infty)\). This behavior defines the motion direction for each interval.

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.

Equilibria
In autonomous differential equations, an equilibrium refers to a state where the system does not change over time. This occurs when the derivative of the function, also known as the rate of change, equals zero. Specifically, for a given differential equation of the form \( y' = f(y) \), equilibria are found by setting \( y' = 0 \), which simplifies to solving \( f(y) = 0 \). Here are the solutions for the provided equations:
  • For \( y' = y^2 - 2 \), we find equilibria by solving \( y^2 - 2 = 0 \), leading to solutions \( y = \pm \sqrt{2} \).
  • For \( y' = \frac{y-3}{y+9} \), equilibria occur when the numerator is zero: \( y - 3 = 0 \), hence \( y = 3 \).
  • For \( y' = y(3-y)(25-y^2) \), equilibria are determined by setting each factor to zero. This results in \( y = 0 \), \( y = 3 \), \( y = 5 \), and \( y = -5 \).
Recognizing these equilibrium points helps in understanding the stationary behavior of a system, which forms the basis for analyzing the dynamic behavior around these points.
Phase Plot
A phase plot graphically represents the equilibrium points and the behavior of trajectories in autonomous differential equations. It provides a visual way to study how solutions evolve over time, identifying stable, unstable, and semi-stable equilibria. Let's consider each equation:
  • For \( y' = y^2 - 2 \), the phase plot examines intervals determined by the roots \( y = \pm \sqrt{2} \). We check the sign of the derivative in the intervals \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, \infty)\). The direction of the arrows in these intervals indicates where the function is increasing or decreasing: positive in the outer intervals and negative between the roots.
  • In the case of \( y' = \frac{y-3}{y+9} \), within \( y \geq 0 \), you examine points around \( y = 3 \). The derivative is negative for \( y < 3 \) and positive for \( y > 3 \), suggesting a stable equilibrium at this point.
  • For \( y' = y(3-y)(25-y^2) \), you analyze multiple intervals, such as \((-\infty, -5)\), \((-5, 0)\), \((0, 3)\), \((3, 5)\), and \((5, \infty)\). Whether \( y' \) is positive or negative in each interval indicates the direction of movement: back and forth across equilibria like a pendulum.
The phase plot is instrumental in visualizing the stability of these equilibria and understanding the transitions between different states.
Analysis of Differential Equations
Analyzing autonomous differential equations involves understanding how solutions behave over time, particularly around equilibrium points. It requires not only finding the equilibria but also evaluating the qualitative behavior of solutions.
  • **Stability:** Equilibrium stability is often assessed through a phase plot. By analyzing how the sign of \( y' \) changes around equilibria, you can determine whether a point is stable (trajectories move towards it), unstable (trajectories move away), or semi-stable (some trajectories approach while others diverge).
  • **Behavior of Solutions:** Consider the signs of derivatives in each interval in the context of the phase plot. Positive derivatives indicate growth or expansion away from a point, while negative ones suggest shrinking or convergence towards a point. Analyzing these signs helps predict the long-term behavior of solutions.
  • **Real World Applications:** Understanding these principles allows us to predict systems' behaviors, from population dynamics to physical systems. Each equilibrium found provides insight into possible steady states, providing crucial data for modeling real-world phenomena.
Through a clear understanding of equilibria and phase plots, you can effectively analyze and interpret the overall dynamics of autonomous differential equations, making these tools indispensable for mathematicians and scientists alike.

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

Use Euler's method with step size 0.2 to estimate \(y(1),\) where \(y(x)\) is the solution of the initial-value problem \(y^{\prime}=x y-x^{2}, y(0)=1\)

Modeling yeast populations In Section 7.1 we introduced the following differential equation to describe the dynamics of an experimental yeast population: $$\frac{d N}{d t}=(0.55-0.0026 N) N$$ where \(N(t)\) is the population size (in millions of individuals per mL) at time \(t\) (in hours). Sketch the direction field of the differential equation for values of \(N\) between 0 and \(250 .\)

(a) Program a calculator or computer to use Euler's method to compute \(y(1),\) where \(y(x)\) is the solution of the initial-value problem $$\frac{d y}{d x}+3 x^{2} y=6 x^{2} \quad y(0)=3$$ for each of the given step sizes. \(\begin{aligned} \text { (i) } h=1 & \text { (ii) } h=0.1 \\ \text { (iii) } h=0.01 & \text { (iv) } h=0.001 \end{aligned}\) (b) Verify that \(y=2+e^{-x^{3}}\) is the exact solution of the differential equation. (c) Find the errors in using Euler's method to compute \(y(1)\) with the step sizes in part (a). What happens to theerror when the step size is divided by 10\(?\)

The logistic differential equation Suppose that the per capita growth rate of a population of size \(N\) declines linearly from a value of \(r\) when \(N=0\) to a value of 0 when \(N=K\) Show that the differential equation for \(N\) is $$\frac{d N}{d t}=r\left(1-\frac{N}{K}\right) N$$

Fitzhugh-Nagumo equations Consider the followingalternative form of the Fitzhugh-Nagumo equations from Example \(3 :\) \(\frac{d v}{d t}=(v-a)(1-v) v-w \quad \frac{d w}{d t}=\varepsilon(v-w)\) where \(\varepsilon>0\) and \(0
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

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.