/*! 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 16 Use a computer system or graphin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate the stability or instability of each critical point of the given differential equation. $$ \frac{d x}{d t}=\left(x^{2}-4\right)^{3} $$

Short Answer

Expert verified
Critical points \(x=2\) and \(x=-2\) are unstable.

Step by step solution

01

Identify Critical Points

To find the critical points of the differential equation \( \frac{dx}{dt} = (x^2 - 4)^3 \), set the equation equal to zero: \((x^2 - 4)^3 = 0\). Solving this, we get \(x^2 - 4 = 0\). Therefore, the critical points are \(x = 2\) and \(x = -2\).
02

Plot the Slope Field

Using a graphing calculator or computer system like Python with Matplotlib or online tools like Desmos, plot the slope field for the differential equation. The slope at each point depends on the value \((x^2 - 4)^3\). Points near \(x=2\) and \(x=-2\) will have nearly horizontal slopes as these are critical points.
03

Analyze Stability at Critical Points

To determine stability, examine the sign of \((x^2 - 4)^3\). For \(x > 2\) or \(x < -2\), \((x^2 - 4)^3 > 0\), indicating a positive slope, suggesting solutions move away from \(x=2\) or \(x=-2\). For \(-2 < x < 2\), \((x^2 - 4)^3 < 0\), indicating a negative slope, suggesting solutions move towards \(x=2\) and \(x=-2\). Therefore, both \(x=2\) and \(x=-2\) are unstable.

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 91Ó°ÊÓ!

Key Concepts

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

Slope Field
A slope field is a visual representation that helps us understand how a differential equation behaves. It consists of little line segments or arrows, each representing the slope of the solution curve at that point. For the differential equation \( \frac{dx}{dt} = (x^2 - 4)^3 \), a slope field is particularly useful to sketch out potential solution curves even without solving the equation explicitly.

To draw a slope field, we calculate the slope \( (x^2 - 4)^3 \) at various points on a plane. This tells us how a potential solution curve would slope at each of those points. Areas where the slope is zero, such as at the critical points, are particularly important because they may be points where the solution curve changes behavior fundamentally.
  • At \( x = 2 \) and \( x = -2 \), the slope is zero, indicating potential critical points where the dynamics may change, making these spots of interest in further analysis.
  • Computational tools like graphing calculators or programming libraries can automate this process, providing a quick and accurate visual representation.
Critical Points
Critical points occur where the derivative of the function (in this case the differential equation) is zero. These are points where the behavior of the system might change, such as switching from increasing to decreasing, or vice versa. For the equation \( \frac{dx}{dt} = (x^2 - 4)^3 \), we find the critical points by solving \( (x^2 - 4)^3 = 0 \).

By factoring the inside of the expression, we get \( x^2 - 4 = 0 \), leading to \( x = 2 \) and \( x = -2 \).
  • These points indicate where the slope changes sign in the slope field. This means they have significant implications for the behavior of solutions near these points.
  • Analyzing these points is crucial as it helps us determine if they are stable or unstable, guiding the understanding of long-term behaviors of solutions.
Stability Analysis
Stability analysis involves determining whether solutions tend to remain near a critical point (stable) or move away as time increases (unstable). After identifying the critical points at \( x = 2 \) and \( x = -2 \), we investigate their stability by checking the sign of \( (x^2 - 4)^3 \) in regions around these points.

  • If \( (x^2 - 4)^3 < 0 \) in a particular region, the slope is negative, indicating that solutions tend to move towards the critical points, suggesting stability.
  • Conversely, if \( (x^2 - 4)^3 > 0 \), solutions move away from the critical points, indicating instability.
For our differential equation:
  • For \( x > 2 \) or \( x < -2 \), \( (x^2 - 4)^3 > 0 \), suggesting instability at these points as solutions diverge away over time.
  • For \( -2 < x < 2 \), \( (x^2 - 4)^3 < 0 \), suggesting a negative slope leading solutions into the critical points. However, this doesn't make \( x = 2 \) and \( x = -2 \) stable but highlights regions of different behavior.
Such analysis is indispensable for predicting the behavior of complex systems modeled by differential equations. Always remember, the goal is to understand the long-term trajectory of solutions once these critical points and their properties are identified.

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

As in Problem 25 of Section \(2.4\), you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem $$ \frac{d v}{d t}=32-1.6 v, \quad v(0)=0 $$ (with \(t\) in seconds and \(v\) in \(\mathrm{ft} / \mathrm{s}\) ). Use the improved Euler method with a programmable calculator or computer to approximate the solution for \(0 \leqq t \leqq 2\), first with step size \(h=0.01\) and then with \(h=0.005\), rounding off approximate \(v\) -values to three decimal places. What percentage of the limiting velocity \(20 \mathrm{ft} / \mathrm{s}\) has been attained after 1 second? After 2 seconds?

Assume that a body moving with velocity \(v\) encounters resistance of the form \(d v / d t=-k v^{3 / 2}\). Show that $$ v(t)=\frac{4 v_{0}}{\left(k t \sqrt{v_{0}}+2\right)^{2}} $$ and that $$ x(t)=x_{0}+\frac{2}{k} \sqrt{v_{0}}\left(1-\frac{2}{k t \sqrt{v_{0}}+2}\right) . $$ Conclude that under a \(\frac{3}{2}\) -power resistance a body coasts only a finite distance before coming to a stop.

Suppose that a community contains 15,000 people who are susceptible to Michaud's syndrome, a contagious disease. At time \(t=0\) the number \(N(t)\) of people who have developed Michaud's syndrome is 5000 and is increasing at the rate of 500 per day. Assume that \(N^{\prime}(t)\) is proportional to the product of the numbers of those who have caught the disease and of those who have not. How long will it take for another 5000 people to develop Michaud's syndrome?

Suppose that a body moves through a resisting medium with resistance proportional to its velocity \(v\), so that \(d v / d t=-k v .\) (a) Show that its velocity and position at time \(t\) are given by $$ v(t)=v_{0} e^{-k t} $$ and $$ x(t)=x_{0}+\left(\frac{v_{0}}{k}\right)\left(1-e^{-k t}\right) $$ (b) Conclude that the body travels only a finite distance, and find that distance.

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the Runge-Kutta method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.25 .\) Construct a table showing five-decimal-place values of the approximate solution and actual solution at the points \(x=0.25\) and \(0.5\). $$ y^{\prime}=-2 x y, y(0)=2 ; y(x)=2 e^{-x^{2}} $$

See all solutions

Recommended explanations on Math Textbooks

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.