Problem 4

Question

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

Step-by-Step Solution

Verified

Answer

Stable equilibrium at $ y = 2 $; unstable at $ y = 0 $.

1Step 1: Find Equilibrium Values

Equilibrium values occur where the derivative is zero. Set $ \frac{dy}{dx} = y^2 - 2y $ to zero. Factoring gives $ y(y-2) = 0 $. Thus, the equilibrium values are $ y = 0 $ and $ y = 2 $.

2Step 2: Determine Stability of Equilibrium Points

To determine stability, consider the derivative $ \frac{dy}{dx} = y^2 - 2y $. Evaluate the sign of $ \frac{dy}{dx} $ on intervals formed by the equilibrium points. For $ y < 0 $, $ \frac{dy}{dx} > 0 $. For $ 0 < y < 2 $, $ \frac{dy}{dx} < 0 $. For $ y > 2 $, $ \frac{dy}{dx} > 0 $. Thus, $ y = 0 $ is an unstable equilibrium and $ y = 2 $ is stable.

3Step 3: Construct a Phase Line

On the phase line, mark the equilibrium points $ y = 0 $ and $ y = 2 $. Draw arrows: on $ (-\infty, 0) $ point right (increasing), on $ (0, 2) $ point left (decreasing), and on $ (2, \infty) $ point right (increasing). This shows that solutions are moving away from $ y = 0 $ and toward $ y = 2 $.

4Step 4: Identify Signs of $y'$ and $y''$

From Step 2, $ y' > 0 $ when $ y < 0 $ and $ y > 2 $, and $ y' < 0 $ when $ 0 < y < 2 $. Compute $ y'' = 2y - 2 $; $ y'' = 2(y-1) $. So $ y'' > 0 $ when $ y > 1 $ and $ y'' < 0 $ when $ y < 1 $.

5Step 5: Sketch Solution Curves

Sketch a graph with several solutions using the information from the phase line. For a curve starting below 0, it should move upwards. Between 0 and 2, curves should move downwards toward $ y = 2 $, and above 2, curves should move downwards toward $ y = 2 $.

Key Concepts

Equilibrium ValuesStability AnalysisPhase Lines

Equilibrium Values

In differential equations, equilibrium values are the points where the derivative of the function equals zero. Essentially, these are the points where the system does not change. To identify these values, you need to set the given derivative equation to zero. For the equation $ \frac{dy}{dx} = y^2 - 2y $, we set it to zero: $ y(y-2) = 0 $. This factoring yields two possible solutions for $ y $, which are $ y = 0 $ and $ y = 2 $.
These values mean that if a system reaches these points, it will remain there unless disturbed by external factors. In this problem, these equilibrium values help us to analyze the stability and behavior of the system.

Stability Analysis

Stability analysis helps to determine whether an equilibrium value is stable or unstable. This analysis involves checking the sign of the derivative $ \frac{dy}{dx} $ in the intervals around each equilibrium point. For $ \frac{dy}{dx} = y^2 - 2y $, checking the intervals:

For $ y < 0 $, $ \frac{dy}{dx} > 0 $, indicating that values are increasing as $ y $ approaches 0. This suggests that $ y = 0 $ is unstable, as solutions tend to move away from this equilibrium.
For $ 0 < y < 2 $, $ \frac{dy}{dx} < 0 $, which means solutions are decreasing, moving towards $ y = 2 $. This marks $ y = 2 $ as stable, since solutions settle around this point.
For $ y > 2 $, $ \frac{dy}{dx} > 0 $, demonstrating that values are increasing again, keeping $ y = 2 $ attractive from above.

This analysis reveals the behavior of the system around each equilibrium point, helping to predict the system's long-term behavior.

Phase Lines

Phase lines visually represent the behavior of solutions to differential equations. Constructing a phase line is a method to combine equilibrium points and their stability into a single diagram. For our function $ \frac{dy}{dx} = y^2 - 2y $, we previously identified equilibrium points and their stability. The creation of a phase line follows:
Place the equilibrium values, $ y = 0 $ and $ y = 2 $, on a number line. Below and above these points, derive the direction of the solutions from the stability analysis:

Arrows pointing right ($ \rightarrow $) from $(-\infty, 0)$ and $(2, \infty)$ indicate increasing solutions.
Arrows pointing left ($ \leftarrow $) between $(0, 2)$ show decreasing tendencies.

This visual system helps in understanding how solutions progress and converge towards or diverge from equilibrium points over time, making these behaviors clearer and intuitive.

Problem 3

Problem 4

Other exercises in this chapter

Problem 3

In Exercises $1-8$ \begin{equation}\begin{array}{l}{\text { a. Identify the equilibrium values. Which are stable and which }} \\ {\text { are unstable? }} \\

View solution

Problem 3

Solve the differential equations in Exercises $1-14$ $$x y^{\prime}+3 y=\frac{\sin x}{x^{2}}, \quad x>0$$

View solution

Problem 4

Solve the differential equations in Exercises $1-14$ $$y^{\prime}+(\tan x) y=\cos ^{2} x, \quad-\pi / 2

View solution

Problem 5

Consider another competitive-hunter model defined by $$ \begin{aligned} \frac{d x}{d t} &=a\left(1-\frac{x}{k_{1}}\right) x-b x y \\\ \frac{d y}{d t} &=m\left(1

View solution