/*! 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 4 In Exercises \(1-8\) \begin{eq... [FREE SOLUTION] | 91Ó°ÊÓ

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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} $$\frac{d y}{d x}=y^{2}-2 y$$

Short Answer

Expert verified
Stable equilibrium at \( y = 2 \); unstable at \( y = 0 \).

Step by step solution

01

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 \( y = 2 \).
02

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.
03

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 \( y = 2 \).
04

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 \( y < 1 \).
05

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 \( y = 2 \).

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Key Concepts

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

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.

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Most popular questions from this chapter

Consider another competitive-hunter model defined by $$ \begin{aligned} \frac{d x}{d t} &=a\left(1-\frac{x}{k_{1}}\right) x-b x y \\\ \frac{d y}{d t} &=m\left(1-\frac{y}{k_{2}}\right) y-n x y \end{aligned} $$ where \(x\) and \(y\) represent trout and bass populations, respectively. $$ \begin{array}{l}{\text { a. What assumptions are implicitly being made about the }} \\ {\text { growth of trout and bass in the absence of competition? }} \\\ {\text { b. Interpret the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { in terms of the }} \\ {\text { physical problem. }}\\\\{\text { c. Perform a graphical analysis: }} \\ {\text { i) Find the possible equilibrium levels. }} \\ {\text { ii) Determine whether coexistence is possible. }} \\ {\text { iii) Pick several typical starting points and sketch typical }} \\ {\text { trajectories in the phase plane. }} \\ {\text { iv) Interpret the outcomes predicted by your graphical }} \\ {\text { analysis in terms of the constants } a, b, m, n, k_{1}, \text { and } k_{2} \text { . }}\end{array} $$ Note: When you get to part (ii), you should realize that five cases exist. You will need to analyze all five cases.

Solve the initial value problems in Exercises \(15-20\) $$(x+1) \frac{d y}{d x}-2\left(x^{2}+x\right) y=\frac{e^{x^{2}}}{x+1}, \quad x>-1, \quad y(0)=5$$

Obtain a slope field and add to it graphs of the solution curves passing through the given points. \(y^{\prime}=y^{2}\) with a. \((0,1) \quad\) b. \((0,2)\) \(\quad\) c. \((0,-1) \quad\) d. \((0,0)\)

Obtain a slope field and add to it graphs of the solution curves passing through the given points. \(y^{\prime}=2(y-4)\) with a. \((0,1) \quad\) b. \((0,4) \quad\) c. \((0,5)\)

Obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. A logistic equation \(y^{\prime}=y(2-y), y(0)=1 / 2 ; 0 \leq x \leq 4,\) \(0 \leq y \leq 3\)

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