/*! 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 27 Find the critical points and pha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 27

Find the critical points and phase portrait of the given autonomous first- order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the \(x y\) -plane determined by the graphs of the equilibrium solutions. $$ \frac{d y}{d x}=y \ln (y+2) $$

Short Answer

Expert verified
Critical points are \( y = 0 \) (unstable) and \( y = -1 \) (asymptotically stable).

Step by step solution

01

Identify Critical Points

To find the critical points of the differential equation \( \frac{dy}{dx} = y \ln(y+2) \), we set \( \frac{dy}{dx} = 0 \). This gives us the equation \( y \ln(y+2) = 0 \). For this product to be zero, either \( y = 0 \) or \( \ln(y+2) = 0 \) must be true.
02

Solve for Critical Points

First, consider \( y = 0 \) as a critical point. Next, solve \( \ln(y+2) = 0 \), which implies \( y+2 = 1 \), giving us \( y = -1 \). Thus, the critical points are \( y = 0 \) and \( y = -1 \).
03

Analyze Stability of Critical Points

To analyze stability, consider the sign of \( \frac{dy}{dx} = y \ln(y+2) \) around each critical point. For \( y = 0 \), if \( y > 0 \), then \( y \ln(y+2) > 0 \), hence solution curves move away from the critical point, indicating that \( y = 0 \) is unstable. For \( y < 0 \) and near \( y = 0 \), \( \ln(y+2) \) becomes negative and \( y \) is negative, making \( y \ln(y+2) > 0 \), also unstable. For \( y = -1 \), if \( y > -1 \), \( y \) will be negative but small, and \( \ln(y+2) \) negative, causing \( y \ln(y+2) < 0 \), suggesting solutions move toward \( y = -1 \). If \( y < -1 \), both \( y \) and \( \ln(y+2) \) are negative, causing \( y \ln(y+2) > 0 \), suggesting solutions move toward \( y = -1 \). Thus, \( y = -1 \) is asymptotically stable.
04

Draw Phase Portrait

In the phase portrait, draw horizontal lines at \( y = 0 \) and \( y = -1 \) representing the critical points. Use arrows to show that solutions diverge from \( y = 0 \) (unstable) and converge to \( y = -1 \) (asymptotically stable). Sample solution curves can be sketched showing the trajectory behavior: moving away from \( y = 0 \) and moving toward and stabilizing at \( y = -1 \).

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.

Critical Points
In the study of differential equations, finding critical points is a crucial step. Critical points occur where the derivative is zero, indicating that the function doesn鈥檛 change at that particular point. In our example, we solve the equation \( \frac{dy}{dx} = y \ln(y+2) = 0 \). The product is zero if either \( y = 0 \) or \( \ln(y+2) = 0 \). Thus, for the equation \( \ln(y+2) = 0 \), we solve for \( y \) and find that \( y = -1 \). We conclude that the critical points are \( y = 0 \) and \( y = -1 \). These points are where the solution does not change, crucial for further stability analysis.
Phase Portrait
A phase portrait provides a visual depiction of solution behaviors over time. Even without computing exact trajectories, we can map out the nature of solutions around critical points. In the depicted phase portrait for our differential equation:
  • We highlight the equilibrium solutions with horizontal lines at \( y = 0 \) and \( y = -1 \).
  • With arrows, movements are shown where solutions diverge from \( y = 0 \) and converge at \( y = -1 \).
  • A typical sketch shows trajectory behavior鈥攈ow solutions move away from or toward critical points.
This visualization helps in understanding the behavior of solutions and identifying potential long-term behaviors.
Stability Analysis
Stability analysis involves determining how small changes in the system can affect the evolution of the system. For stability, we analyze what happens to solutions as they approach or move away from the critical points.
  • Unstable Point: For \( y = 0 \), if \( y > 0 \) or \( y < 0 \), \( \ln(y+2) \) ensures \( y \ln(y+2) > 0 \), indicating solutions are diverging, classifying \( y = 0 \) as unstable.
  • Asymptotically Stable Point: At \( y = -1 \), solutions behave differently: when \( y > -1 \) or \( y < -1 \), calculations show that solutions converge, classifying \( y = -1 \) as asymptotically stable.
Stability analysis helps predict how systems behave over time and which points may potentially attract solutions.
Equilibrium Solutions
Equilibrium solutions are the constant solutions of a differential equation, providing insight into the behavior of systems over time. For \( \frac{dy}{dx} = y \ln(y+2) \), we previously found our critical points to be \( y = 0 \) and \( y = -1 \). These are not only critical points but also equilibrium solutions where the rate of change of \( y \) is zero. At these points, the system remains in its current state:
  • At \( y = 0 \): Although it鈥檚 an equilibrium point, it's unstable. Any minor change causes the solutions to deviate.
  • At \( y = -1 \): It is a stable equilibrium where solutions tend to settle down, signifying reliable equilibrium states.
Understanding equilibrium solutions is vital to comprehend how dynamical systems maintain or change their states under varying conditions.

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

Consider the competition model defined by $$ \begin{aligned} &\frac{d x}{d t}=x(2-0.4 x-0.3 y) \\ &\frac{d y}{d t}=y(1-0.1 y-0.3 x) \end{aligned} $$ where the populations \(x(t)\) and \(y(t)\) are measured in the thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the cases: (a) \(x(0)=1.5, \quad y(0)=3.5\) (b) \(x(0)=1, \quad y(0)=1\) (c) \(x(0)=2\) \(y(0)=7\) (d) \(x(0)=4.5\), \(y(0)=0.5\)

Devise an appropriate substitution to solve $$ x y^{\prime}=y \ln (x y) $$

Suppose an \(R C\) -series circuit has a variable resistor. If the resistance at time \(t\) is given by \(R=k_{1}+k_{2} t\), where \(k_{1}\) and \(k_{2}\) are known positive constants, then (9) becomes $$ \left(k_{1}+k_{2} t\right) \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ If \(E(t)=E_{0}\) and \(q(0)=q_{0}\), where \(E_{0}\) and \(q_{0}\) are constants, show that $$ q(t)=E_{0} C+\left(q_{0}-E_{0} C\right)\left(\frac{k_{1}}{k_{1}+k_{2} t}\right)^{1 / C k_{2}} $$

Use a numerical solver to obtain a numerical solution curve for the given initial-value problem. First use Euler's method and then the RK4 method. Use \(h=0.25\) in each case. Superimpose both solution curves on the same coordinate axes. If possible, use a different color for each curve. Repeat, using \(h=0.1\) and \(h=0.05\). $$ y^{\prime}=y(10-2 y), \quad y(0)=1 $$

The differential equation $$ \frac{d y}{d x}=P(x)+Q(x) y+R(x) y^{2} $$ is known as Riccati's equation. (a) A Riccati equation can be solved by a succession of two substitutions provided we know a particular solution \(y_{1}\) of the equation. Show that the substitution \(y=y_{1}+u\) reduces Riccati's equation to a Bernoulli equation (4) with \(n=2\). The Bernoulli equation can then be reduced to a linear equation by the substitution \(w=u^{-1}\).
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Waves Physics

Read Explanation

Electricity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Kinematics Physics

Read Explanation

Physics of Motion

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.