/*! 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 17 In Problems, use a computer syst... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems, 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. (Some of these critical points may be semistable in the sense mentioned in Example 6.) $\$$ $$ \frac{d x}{d t}=x^{2}\left(x^{2}-4\right) $$

Short Answer

Expert verified
Critical points: \( x = 0, 2, -2 \); \( x = 0 \) is semistable, \( x = -2 \) is stable, and \( x = 2 \) is unstable.

Step by step solution

01

Identify the Differential Equation

The given differential equation is \( \frac{dx}{dt} = x^2(x^2 - 4) \). We will use this equation to find the critical points and determine their stability or instability.
02

Find Critical Points

Critical points occur where \( \frac{dx}{dt} = 0 \). Set the equation \( x^2(x^2 - 4) = 0 \). This implies \( x^2 = 0 \) or \( x^2 - 4 = 0 \). Solving these gives critical points at \( x = 0 \), \( x = 2 \), and \( x = -2 \).
03

Analyze Stability

To analyze stability, examine the factorization: \( \frac{dx}{dt} = x^2(x - 2)(x + 2) \). Consider the sign of \( \frac{dx}{dt} \) around each critical point: - **Around \(x = 0\):** - For \(x < 0\), \(x^2 > 0\).\( x+2 < 0\), \(x-2 < 0\) so \( \frac{dx}{dt} < 0 \). - For \(x > 0\), \(x^2 > 0\).\( x+2 > 0\),\(x-2>0\), so \( \frac{dx}{dt} > 0 \).Thus, \( x=0 \) is a semistable (unstable from below, stable from above).- **Around \(x = -2\):** - For \(x < -2\), \(x + 2 < 0\),\(x - 2 < 0\), so \( \frac{dx}{dt} > 0 \). - For \(-2 < x < 0\), \(x^2 > 0\),\(x+2 > 0\), \(x-2 < 0\), so \( \frac{dx}{dt} < 0 \).Thus, \( x = -2 \) is stable.- **Around \(x = 2\):** - For \(x < 2\), \(x + 2 > 0\),\(x - 2 > 0\), so \( \frac{dx}{dt} < 0 \). - For \(x > 2\), \(x + 2 > 0\),\(x - 2 > 0\), so \( \frac{dx}{dt} > 0 \).Thus, \( x = 2 \) is unstable.
04

Verify with Plotting

Use a graphing tool or software to plot the slope field of the differential equation. Observe the behavior of trajectories around \( x = 0 \), \( x = -2 \), and \( x = 2 \). - Verify semistability at \( x = 0 \): trajectories move away downward but stabilize from above.- Verify stability at \( x = -2 \): trajectories stabilize at \( x = -2 \). - Verify instability at \( x = 2 \): trajectories move away upwards and downwards.

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Key Concepts

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

Critical Points
Critical points are the values of the variable, in this case, "x," where the derivative of the differential equation equals zero. For the equation \( \frac{dx}{dt} = x^2(x^2 - 4) \), we set the derivative to zero: \( x^2(x^2 - 4) = 0 \). This indicates that the factor \( x^2 \) or \( x^2 - 4 \) must be zero. Solving for \( x^2 = 0 \) gives us \( x = 0 \). Solving \( x^2 - 4 = 0 \), we have \( x = 2 \) and \( x = -2 \). So, we have critical points at \( x = 0 \), \( x = 2 \), and \( x = -2 \). These points are where the behavior of the solution changes, meaning they are either stable, unstable, or semistable. Identifying these points is a crucial step in understanding the overall behavior of the system.
Stability Analysis
In stability analysis, we determine whether small disturbances at critical points diverge or converge. This helps us understand if the solutions will stabilize or not. We can check this by analyzing the sign of the derivative \( \frac{dx}{dt} \) around each critical point. For example, around \( x = 0 \):- If \( x < 0 \), \( \frac{dx}{dt} < 0 \). - If \( x > 0 \), \( \frac{dx}{dt} > 0 \). This indicates that \( x = 0 \) is semistable, meaning solutions are unstable from below but stable from above. Around \( x = -2 \): - If \( x < -2 \), \( \frac{dx}{dt} > 0 \). - If \( -2 < x < 0 \), \( \frac{dx}{dt} < 0 \). So, \( x = -2 \) is stable as the solutions converge.For \( x = 2 \): - If \( x < 2 \), \( \frac{dx}{dt} < 0 \). - If \( x > 2 \), \( \frac{dx}{dt} > 0 \). Here, trajectories move away from the point, indicating instability at \( x = 2 \). Understanding this behavior for each point helps you to predict how solutions behave in the neighborhood of these critical points.
Slope Field
A slope field provides a visual representation of the direction of solutions to a differential equation at any given point on the plane. It consists of small line segments or arrows indicating the slope of the solution at selected points. To create a slope field for the equation \( \frac{dx}{dt} = x^2(x^2 - 4) \), you would plot arrows on a graph where the direction and steepness of each arrow correspond to the derivative's value \( \frac{dx}{dt} \) at that particular point \( x \). The slope field helps you visualize how different initial conditions lead to different trajectories in the solution. This graphical tool allows you to see the tendencies of the solutions to diverge or converge around critical points. By observing the slope field, you can confirm predictions made in the analytical stability analysis of the critical points.
Solution Curves
Solution curves are trajectories that represent the solutions or paths followed by a differential equation over time. When plotted on a slope field, these curves provide an insight into the long-term behavior of solutions starting from various initial conditions. To find the solution curves, select initial points and trace the path, following the arrows or line segments in the slope field. - For \( x = 0 \), the solution curve shows semistable behavior.- At \( x = -2 \), the solution curves converge, indicating stability. - Around \( x = 2 \), the solution curves show divergence, signaling instability.By observing these solution curves, especially how they behave near the critical points, you can draw conclusions about the stability of each critical point and the dynamic nature of the system described by the differential equation.

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Most popular questions from this chapter

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. Suppose that the fish population \(P(t)\) in a lake is attacked by a disease at time \(t=0\), with the result that the fish cease to reproduce (so that the birth rate is \(\beta=0\) ) and the death rate \(\delta\) (deaths per week per fish) is thereafter proportional to \(1 / \sqrt{P}\). If there were initially 900 fish in the lake and 441 were left after 6 weeks, how long did it take all the fish in the lake to die?

In Jules Verne's original problem, the projectile launched from the surface of the earth is attracted by both the earth and the moon, so its distance \(r(t)\) from the center of the earth satisfies the initial value problem $$ \frac{d^{2} r}{d t^{2}}=-\frac{G M_{e}}{r^{2}}+\frac{G M_{m}}{(S-r)^{2}} ; \quad r(0)=R, \quad r^{\prime}(0)=v_{0} $$ where \(M_{e}\) and \(M_{m}\) denote the masses of the earth and the moon, respectively; \(R\) is the radius of the earth and \(S=384,400 \mathrm{~km}\) is the distance between the centers of the earth and the moon. To reach the moon, the projectile must only just pass the point between the moon and earth where its net acceleration vanishes. Thereafter it is "under the control" of the moon, and falls from there to the lunar surface. Find the minimal launch velocity \(v_{0}\) that suffices for the projectile to make it "From the Earth to the Moon."

The mass of the sun is 329,320 times that of the earth and its radius is 109 times the radius of the earth. (a) To what radius (in meters) would the earth have to be compressed in order for it to become a black hole - the escape velocity from its surface equal to the velocity \(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\) of light? (b) Repeat part (a) with the sun in place of the earth.

Example 4 dealt with the case \(4 h>k M^{2}\) in the equation \(d x / d t=k x(M-x)-h\) that describes constant-rate harvesting of a logistic population. Problems deal with the other cases. If \(4 h>k M^{2}\), show that \(x(t)=0\) after a finite period of time, so the lake is fished out (whatever the initial population). [Suggestion: Complete the square to rewrite the differential equation in the form \(d x / d t=-k\left[(x-a)^{2}+b^{2}\right]\). Then solve explicitly by separation of variables.] The results of this and the previous problem (together with Example 4) show that \(h=\frac{1}{4} k M^{2}\) is a critical harvesting rate for a logistic population. At any lesser harvesting rate the population approaches a limiting population \(N\) that is less than \(M\) (why?), whereas at any greater harvesting rate the population reaches extinction.

A computer with a printer is required for Problems 17 through 24\. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to \(x\). \(y^{\prime}=x+\sqrt{y}, y(0)=1 ; 0 \leqq x \leqq 2\)

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