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Chapter 9: Problem 4

In Exercises \(1-8\) a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$ \frac{d y}{d x}=y^{2}-2 y $$

Short Answer

Expert verified
Equilibrium points are at \( y = 0 \) (stable) and \( y = 2 \) (unstable).

Step by step solution

01

Identify Equilibrium Values

To find equilibrium points, set \( \frac{dy}{dx} = 0 \). Thus, the equation becomes \( y^2 - 2y = 0 \). Factor the equation: \( y(y - 2) = 0 \). Therefore, the equilibrium values are \( y = 0 \) and \( y = 2 \).
02

Determine Stability of Equilibrium Points

Evaluate the derivative \( \frac{dy}{dx} = y(y - 2) \). At the equilibrium point \( y = 0 \), choose a test point: e.g., \( y = 1 \), then \( \frac{dy}{dx} = 1(1 - 2) = -1 \), indicating \( y = 0 \) is stable. For \( y = 2 \), choose a test point: e.g., \( y = 3 \), then \( \frac{dy}{dx} = 3(3 - 2) = 3 \), indicating \( y = 2 \) is unstable.
03

Construct the Phase Line

Draw a vertical line representing \( y \). Mark the equilibrium points \( y = 0 \) and \( y = 2 \). Above \( y = 2 \), \( \frac{dy}{dx} > 0 \), indicating arrows pointing upwards. Between \( y = 0 \) and \( y = 2 \), \( \frac{dy}{dx} < 0 \), indicating arrows pointing downwards. Below \( y = 0 \), \( \frac{dy}{dx} > 0 \), also pointing upwards.
04

Analyze Signs of \( y' \) and \( y'' \)

Take the derivative of \( \frac{dy}{dx} = y^2 - 2y \) to find \( y'' \): \( y'' = \frac{d}{dy}(y^2 - 2y) = 2y - 2 \). Analyze this derivative for intervals defined in Step 3. For \( y < 0 \), \( y'' < 0 \); for \( 0 < y < 2 \), \( y'' < 0 \); for \( y > 2 \), \( y'' > 0 \).
05

Sketch the Solution Curves

Plot the points and arrows indicating the direction and concavity along the vertical axis based on the phase line and stability analysis. Draw solution curves approaching the stable equilibrium at \( y = 0 \) and curving away from the unstable equilibrium at \( y = 2 \). The curves demonstrate decreased gradient between equilibria and increased gradient away from \( y = 2 \).

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Key Concepts

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

Equilibrium Values
Equilibrium values, also known as fixed points, are crucial in understanding differential equations. These values occur when the rate of change is zero, meaning the value of the function remains constant over time at these points.

For the differential equation \( \frac{dy}{dx} = y^2 - 2y \), we find equilibrium values by setting this equation equal to zero: \( y^2 - 2y = 0 \). By factoring, \( y(y - 2) = 0 \), we identify the equilibrium values as \( y = 0 \) and \( y = 2 \).

These values tell us where the function doesn't change as time progresses, which allows us to understand the potential long-term behaviors of the system.

  • **Stable equilibrium**: This is a point where if the system is disturbed, it will return to the equilibrium point.
  • **Unstable equilibrium**: Here, if the system is disturbed, it will move away from the equilibrium point.
Phase Line
The phase line is a visual tool that helps to understand the behavior of solutions to differential equations. It is a vertical line on which the equilibrium values are marked.

To construct a phase line for \( \frac{dy}{dx} = y^2 - 2y \):
  • Draw a vertical line representing \( y \).
  • Mark the equilibrium points \( y = 0 \) and \( y = 2 \).
  • Based on the derivative's sign, draw arrows indicating the direction of change:

- Below \( y = 0 \), since \( \frac{dy}{dx} > 0 \), draw arrows pointing upwards.
- Between \( y = 0 \) and \( y = 2 \), since \( \frac{dy}{dx} < 0 \), draw arrows pointing downwards.
- Above \( y = 2 \), since \( \frac{dy}{dx} > 0 \), draw arrows pointing upwards.

This line helps to quickly visualize where the function is increasing or decreasing, providing insights into the system's dynamics.
Stability Analysis
Stability analysis allows us to understand whether an equilibrium point is stable or unstable. We analyze the behavior of the system near these points.

In the given differential equation \( \frac{dy}{dx} = y^2 - 2y \):
  • To assess stability, consider test points near equilibriums.
  • For \( y = 0 \), choose a test point such as \( y = 1 \): we find \( \frac{dy}{dx} = -1 \), indicating stability since the solution returns.
  • For \( y = 2 \), test a point like \( y = 3 \): we find \( \frac{dy}{dx} = 3 \), indicating instability since the solution moves away.

This technique shows which values make solutions diverge or converge, guiding predictions about how the system behaves over time.
Solution Curves
Solution curves help in visualizing how the solutions behave over different regions on the phase line. They provide a complete picture of how the system evolves.

To sketch the solution curves for the differential equation \( \frac{dy}{dx} = y^2 - 2y \):
  • Observe how solution curves approach the stable point \( y = 0 \).
  • Notice that solutions curve away from the unstable point \( y = 2 \).
  • The slope indications from earlier steps guide the concavity and direction.

The curves illustrate the dynamics of stability, showing the rapidity at which systems reach or move away from equilibrium points. Such visual tools enrich the understanding of differential equation behavior in real-world scenarios.

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Most popular questions from this chapter

Is either of the following equations correct? Give reasons for your answers. a. \(x \int \frac{1}{x} d x=x \ln |x|+C\) b. \(x \int \frac{1}{x} d x=x \ln |x|+C x\)

Suppose \(a\) and \(b\) are positive numbers. Sketch the parabolas $$ y^{2}=4 a^{2}-4 a x \quad \text { and } \quad y^{2}=4 b^{2}+4 b x $$ in the same diagram. Show that they intersect at \((a-b, \pm 2 \sqrt{a b})\) and that each " \(a\) -parabola"" is orthogonal to every "b-parabola."

Use the Euler method with \(d x=0.2\) to estimate \(y(1)\) if \(y^{\prime}=y\) and \(y(0)=1 .\) What is the exact value of \(y(1) ?\)

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y-\) window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b] .\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {0 \leq x \leq 4, \quad 0 \leq y \leq 3 ; \quad b=3}\end{array} $$

Solve the differential equations. \(\sin \theta \frac{d r}{d \theta}+(\cos \theta) r=\tan \theta, \quad 0<\theta<\pi / 2\)
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