/*! 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 25 Solving \(y^{2}\left(4-y^{2}\rig... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 25

Solving \(y^{2}\left(4-y^{2}\right)=y^{2}(2-y)(2+y)=0\) we obtain the critical points \(-2,0,\) and \(2 .\) From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is semi-stable, and -2 is unstable (repeller).

Short Answer

Expert verified
Critical points are -2, 0, and 2 with respective stabilities: repeller, semi-stable, and attractor.

Step by step solution

01

Set each factor to zero

Given the equation \(y^2(4 - y^2) = y^2(2 - y)(2 + y) = 0\), we start by setting each factor equal to zero to find the solutions. This gives us: \(y^2 = 0\), \(2 - y = 0\), and \(2 + y = 0\).
02

Solve each equation

For \(y^2 = 0\), solving gives \(y = 0\).For \(2 - y = 0\), solving gives \(y = 2\).For \(2 + y = 0\), solving gives \(y = -2\).
03

Identify critical points

The solutions \(y = -2\), \(y = 0\), and \(y = 2\) are the critical points of the system. These points denote where the system changes behavior.
04

Analyze stability from phase portrait

Referencing the phase portrait analysis given, we determine the stability of each critical point:- \(y = 2\) is an attractor, indicating asymptotic stability.- \(y = 0\) is semi-stable, meaning it is stable on one side and unstable on the other.- \(y = -2\) is a repeller, indicating instability.

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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 specific values of a variable where the behavior of a differential equation changes. This is determined when the derivative or the output of the system equals zero. In our case, solving the equation \(y^2(4-y^2) = 0\) provides the critical points \(-2, 0,\) and \(2\). These points capture the essence of the system's dynamics. Each critical point tells us where the system could potentially rest (equilibrium). However, not all critical points are the same, as their stability can differ greatly.
Phase Portrait
A phase portrait is a graphical tool used to visualize the behavior of a dynamical system. By plotting trajectories of solutions in a phase space, it shows how stable or unstable each state may be over time. In our exercise, the phase portrait illustrates how the solutions evolve near the critical points. This visualization helps in understanding whether solutions move towards or away from these points. It aids in predicting the system's long-term behavior by providing a visual cue of motion and changes around those critical points.
Stability Analysis
Stability analysis involves examining whether small disturbances to a system lead to the solution remaining near the critical points (stable) or drifting away (unstable). By analyzing the critical points in our equation, we can determine their stability properties. In our exercise, the analysis categorizes the points: \(y = 2\) as asymptotically stable, \(y = 0\) as semi-stable, and \(y = -2\) as unstable. Stability analysis is crucial for understanding how sensitive a system is to small changes, which reveals important insights into the nature and safety of the system.
Asymptotic Stability
Asymptotic stability refers to a condition where, after a disturbance, the system returns to the critical point over time. It's like a ball rolling back to the bottom of a bowl after being pushed. For the point \(y = 2\), the phase portrait shows that over time, any trajectory starting near \(y = 2\) will eventually approach and settle at this point. Asymptotic stability is a desirable property in many systems, as it indicates robustness and resiliency, ensuring that a system can return to equilibrium after minor disruptions.
Unstable Points
Unstable points are critical points where small disturbances cause the system to move away from the equilibrium, like a ball perched on the top of a hill. In the case of \(y = -2\), a slight deviation leads the system to diverge further away from the critical point. This behavior is shown in the phase portrait, where trajectories divert away from \(-2\). Understanding unstable points is essential because they signify regions of high sensitivity within a system, and they indicate where control measures must be implemented to maintain a system's functionality.

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Most popular questions from this chapter

Separating variables we have \(d y /\left(\sqrt{1+y^{2}} \sin ^{2} y\right)=d x\) which is not readily integrated (even by a CAS). We note that \(d y / d x \geq 0\) for all values of \(x\) and \(y\) and that \(d y / d x=0\) when \(y=0\) and \(y=\pi,\) which are equilibrium solutions.

When the height of the water is \(h,\) the radius of the top of the water is \(\frac{2}{5}(20-h)\) and \(A_{w}=4 \pi(20-h)^{2} / 25 .\) The differential equation is \\[ \frac{d h}{d t}=-c \frac{A_{h}}{A_{w}} \sqrt{2 g h}=-0.6 \frac{\pi(2 / 12)^{2}}{4 \pi(20-h)^{2} / 25} \sqrt{64 h}=-\frac{5}{6} \frac{\sqrt{h}}{(20-h)^{2}} \\] Separating variables and integrating we have \\[ \frac{(20-h)^{2}}{\sqrt{h}} d h=-\frac{5}{6} d t \quad \text { and } \quad 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+c \\] Using \(h(0)=20\) we find \(c=2560 \sqrt{5} / 3,\) so an implicit solution of the initial-value problem is \\[ 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+\frac{2560 \sqrt{5}}{3} \\] To find the time it takes the tank to empty we set \(h=0\) and solve for \(t .\) The tank empties in \(1024 \sqrt{5}\) seconds or 38.16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom.

(a) \(\mathrm{By}\) inspection \(y=x\) and \(y=-x\) are solutions of the differential equation and not members of the family \(y=x \sin \left(\ln x+c_{2}\right).\) (b) Letting \(x=5\) and \(y=0\) in \(\sin ^{-1}(y / x)=\ln x+c_{2}\) we get \(\sin ^{-1} 0=\ln 5+c\) or \(c=-\ln 5 .\) Then \(\sin ^{-1}(y / x)=\ln x-\ln 5=\ln (x / 5) .\) Because the range of the arcsine function is \([-\pi / 2, \pi / 2]\) we must have $$\begin{array}{c} -\frac{\pi}{2} \leq \ln \frac{x}{5} \leq \frac{\pi}{2} \\ e^{-\pi / 2} \leq \frac{x}{5} \leq e^{\pi / 2} \\ 5 e^{-\pi / 2} \leq x \leq 5 e^{\pi / 2}. \end{array}$$ The interval of definition of the solution is approximately [1.04, 24.05].

(a) \(\operatorname{From} d P / d t=\left(k_{1}-k_{2}\right) P\) we obtain \(P=P_{0} e^{\left(k_{1}-k_{2}\right) t}\) where \(P_{0}=P(0)\). (b) If \(k_{1} >k_{2}\) then \(P \rightarrow \infty\) as \(t \rightarrow \infty\). If \(k_{1}=k_{2}\) then \(P=P_{0}\) for every \(t .\) If \(k_{1}< k_{2}\) then \(P \rightarrow 0\) as \(t \rightarrow \infty\).

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).
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