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Chapter 2: Problem 28

Solving \(y e^{y}-9 y=y\left(e^{y}-9\right)=0\) we obtain the critical points 0 and \(\ln 9 .\) From the phase portrait we see that 0 is asymptotically stable (attractor) and \(\ln 9\) is unstable (repeller).

Short Answer

Expert verified
Critical points are 0 (stable) and \( \ln 9 \) (unstable).

Step by step solution

01

Define the Equation

The given equation is \( y(e^y - 9) = 0 \). This product equals zero if and only if either \( y = 0 \) or \( e^y - 9 = 0 \).
02

Solve for the First Critical Point

The equation \( y = 0 \) is straightforward, giving us the first critical point \( y = 0 \).
03

Solve for the Second Critical Point

The equation \( e^y - 9 = 0 \) can be rewritten as \( e^y = 9 \). Taking the natural logarithm on both sides, we get \( y = \ln 9 \), thus, the second critical point is \( y = \ln 9 \).
04

Analyze Stability

The problem states that the phase portrait shows \( y = 0 \) is asymptotically stable (an attractor), while \( y = \ln 9 \) is unstable (a repeller). This aligns with analytical stability checks where the derivative of the function at these points can show the nature of each critical point.

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Key Concepts

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

Critical Points in Differential Equations
Critical points are crucial in understanding the behavior of differential equations. They are the values of the variable where the derivative (or rate of change) of a function is zero, indicating a potential equilibrium state. In the given problem, we find critical points by setting the equation \( y(e^y - 9) = 0 \) to find where changes stop. We identify two critical points: \( y = 0 \) and \( y = \ln 9 \).
  • For \( y = 0 \), it's clear from directly solving \( y = 0 \).
  • For \( y = \ln 9 \), it results from solving \( e^y = 9 \), using logarithmic properties.
Critical points help determine the overall behavior of a system over time, making them essential for more advanced stability analysis.
Understanding Phase Portraits
Phase portraits are visual representations that illustrate how the state of a system changes over time. In simpler terms, they help us visualize how solutions to a differential equation evolve. In one-dimensional systems, phase portraits typically feature arrows that illustrate the direction of movement across the real line.
  • When a critical point is an attractor, arrows point toward it, indicating stability as nearby points converge to it over time.
  • If a point is a repeller, arrows direct away, showing nearby points diverge, meaning instability.
In our specific example, the phase portrait reveals:- \( y = 0 \) is stable (attracts arrows), marking it as an attractor.- \( y = \ln 9 \) is unstable (repels arrows), acting as a repeller.
Through these visual aids, phase portraits are powerful tools for quickly grasping the stability of critical points.
Basics of Stability Analysis
Stability analysis is essential for predicting the behavior of differential equations near critical points. By examining how solutions change near these points, we determine whether a critical state is stable, unstable, or a saddle point.
To perform this analysis, especially in a simple case:
  • Calculate the derivative of the function at the critical point.
  • Check the sign of the derivative:
    • If positive, the point is unstable (repeller).
    • If negative, it's stable (attractor).
    • If zero, further analysis is needed, often indicating a more complex behavior.
For the discussed exercise, \( y = 0 \) shows negative tendencies in nearby regions, while \( y = \ln 9 \) tends to diverge from it, matching the phase portrait observations.
Applying the Natural Logarithm
The natural logarithm is a fundamental concept in solving equations involving exponential functions, such as \( e^y \). It's denoted as \( \ln \), and its base is \( e \) (approximately 2.71828).
  • It transforms multiplicative relationships into additive ones, simplifying equations.
  • Crucial for solving equations like \( e^y = a \), transforming to \( y = \ln a \).
  • Reflects the concept of continuous growth 鈥� important for modeling real-world processes.
In this exercise, the equation \( e^y = 9 \) was simplified using the natural logarithm, resulting in the second critical point, \( y = \ln 9 \).
Understanding the logarithm's role helps simplify complex expressions and discover critical points with ease.

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Most popular questions from this chapter

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

Assume that \(d T / d t=k(T-100)\) so that \(T=100+c e^{k t} .\) If \(T(0)=20^{\circ}\) and \(T(1)=22^{\circ},\) then \(c=-80\) and \(k=\ln (39 / 40)\) so that \(T(t)=90^{\circ},\) which implies \(t=82.1\) seconds. If \(T(t)=98^{\circ}\) then \(t=145.7\) seconds.

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) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integrating, we get \\[\ln |y|-\frac{1}{2} \ln |1-y|-\frac{1}{2} \ln |1+y|=x+c\\]. When \(y>1,\) this becomes \\[\ln y-\frac{1}{2} \ln (y-1)-\frac{1}{2} \ln (y+1)=\ln \frac{y}{\sqrt{y^{2}-1}}=x+c.\\] Letting \(x=0\) and \(y=2\) we find \(c=\ln (2 / \sqrt{3}) .\) Solving for \(y\) we get \(y_{1}(x)=2 e^{x} / \sqrt{4 e^{2 x}-3},\) where \(x>\ln (\sqrt{3} / 2)\) When \(0\ln (\sqrt{3} / 2)\).
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