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Chapter 2: Problem 28

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. $$\frac{d y}{d x}=\frac{y e^{y}-9 y}{e^{y}}$$

Short Answer

Expert verified
The critical points are at \( y = 0 \) (unstable) and \( y = \ln(9) \) (stable).

Step by step solution

01

Simplifying the Differential Equation

Given the differential equation \( \frac{d y}{d x} = \frac{y e^{y} - 9y}{e^{y}} \). Simplify it by factoring out \( y \) from the numerator: \( y \frac{e^{y} - 9}{e^{y}} \). This can be further simplified to \( \frac{d y}{d x} = y \left(1 - \frac{9}{e^{y}}\right) \).
02

Finding Critical Points

Critical points occur where \( \frac{d y}{d x} = 0 \). So set \( y \left(1 - \frac{9}{e^{y}}\right) = 0 \). This yields the conditions: \( y = 0 \) or \( 1 - \frac{9}{e^{y}} = 0 \). Solving \( 1 = \frac{9}{e^{y}} \) gives \( e^{y} = 9 \), thus \( y = \ln(9) \). Critical points are \( y = 0 \) and \( y = \ln(9) \).
03

Analyzing Stability of Critical Points

Analyze the stability by evaluating the derivative of \( f(y) = y \left(1 - \frac{9}{e^{y}}\right) \). The derivative is \( f'(y) = 1 - \frac{9}{e^{y}} + y \left(\frac{9 e^{y}}{e^{2y}}\right). \) Evaluate at \( y = 0 \) and \( y = \ln(9) \) to determine stability. If \( f'(y) < 0 \), the critical point is stable; if \( f'(y) > 0 \), it's unstable.
04

Sketching Phase Portrait

The phase portrait can be visualized on the \( y \)-axis, with arrows indicating the direction of \( \frac{d y}{d x} \). Near \( y = 0 \), \( f(y) = 0 \) and \( f'(y) > 0 \) indicates instability. Near \( y = \ln(9) \), \( f(y) = 0 \) and \( f'(y) < 0 \) suggests stability. Sketch arrows moving away from \( y = 0 \) and towards \( y = \ln(9) \).
05

Conclusion: Classifying Critical Points

The critical point at \( y = 0 \) is unstable as \( f'(y) > 0 \) and solutions move away. The critical point at \( y = \ln(9) \) is asymptotically stable as \( f'(y) < 0 \) and solutions move towards it.

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Key Concepts

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

Critical Points
In the realm of differential equations, critical points play a significant role. They represent values where the rate of change, or derivative, equals zero. In simpler terms, at these points, the slope flattens out, indicating potential equilibrium states where the system might settle. For the exercise at hand, we simplify the differential equation to identify critical points.
  • To find critical points, set the derivative to zero.
  • For the given equation: \( y \left(1 - \frac{9}{e^{y}}\right) = 0 \).
Solving this gives us two critical points: \( y = 0 \) and \( y = \ln(9) \). These points are significant because they help us understand where the system reaches steady states. Recognizing them is the first step in family unraveling the behavior of the equation.
Phase Portrait
Visualizing the behavior of differential equations can be greatly aided by phase portraits. These are graphical representations that display various trajectories solutions may follow in the state space. In our example, since we deal with an autonomous equation, the phase portrait describes how solutions evolve over time without external forcing.
  • In a phase portrait, arrows indicate the direction of solution movement over time, either towards or away from critical points.
  • The axis represents all possible values of \( y \), and we observe how the system's behavior shifts across these values.
By examining the phase portrait for this differential equation, notice how arrows move away from \( y = 0 \) indicating instability, and flow towards \( y = \ln(9) \), showing stability. This graphical insight helps to predict long-term behavior of the solutions.
Stability Analysis
Assessing the stability of critical points is crucial as it determines the nature of these points - whether they attract, repel, or have neutral effects on nearby solutions. Stability analysis involves calculating derivatives and interpreting signs.
  • Evaluate the derivative \( f'(y) \) at critical points to assess stability.
  • If \( f'(y) < 0 \), the point is asymptotically stable, meaning nearby solutions tend to stabilize at this point.
  • If \( f'(y) > 0 \), the point is unstable, causing solutions to move away.
For our exercise, at \( y = 0 \), since \( f'(y) > 0 \), the point is unstable. Conversely, at \( y = \ln(9) \), \( f'(y) < 0 \) signals asymptotic stability. This analysis offers vital information on how the system behaves around these critical points.
Autonomous First-Order Differential Equation
An "autonomous first-order differential equation" is a type of differential equation that does not explicitly depend on the independent variable. This characteristic simplifies analysis since it allows the description of the system's dynamics solely based on the state variable - herein \( y \).
  • These equations are expressed in the form \( \frac{dy}{dx} = f(y) \), highlighting that the rate of change depends only on \( y \) itself.
  • This autonomy grants the system a time-invariant nature, where the behavior remains consistent regardless of when it is observed.
In the given equation, the autonomous nature allows for direct examination of \( y \) and significantly aids the understanding of the underpinning dynamics, facilitating the analysis of critical points and phase portraits. Understanding this fundamental concept is crucial for studying how the solutions of differential equations evolve inherently over time.

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Most popular questions from this chapter

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$y^{\prime}+(\tan x) y=\cos ^{2} x, \quad y(0)=-1$$

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