Location Income Sheet - Urban Credit Balance (2021) ✓ Solved

Sheet1locationincome1000sizeyearscredit Balanceurban543124016rur

Sheet1locationincome1000sizeyearscredit Balanceurban543124016rur

Sheet1 LOCATION INCOME($1000) SIZE YEARS CREDIT BALANCE($) Urban Rural Suburban Suburban Rural Urban Rural Urban Suburban Urban Urban Urban Rural Rural Urban Suburban Urban Urban Rural Urban Suburban Urban Suburban Rural Urban Suburban Rural Rural Urban Suburban Rural Urban Urban Urban Urban Urban Suburban Urban Suburban Rural Urban Suburban Suburban Rural Urban Suburban Rural Rural Suburban Suburban Sheet2 Sheet3 Homework Section 10.5 MTH 122 Name: ________________________________________ Directions: Complete all questions. Your work must be legible, neat, and complete. Show your work and provide explanations where needed. All pages need to be stapled. All graphs need to be drawn neatly and any lines should be drawn using a straight edge.

Questions: One or more initial conditions are given for each differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a yz-graph. Always indicate the constant solutions on the ty-graph whether they are mentioned or not. 1. [3 points] y′ = 23 y − 3 y(0) = 3 y(0) = 6 2. [4 points] y′ = (y − 1)(y − 5) y(0) = −2 y(0) = 2 y(0) = 4 y(0) = 6 3. [3 points] y′ = −y2 + 10y − 21 y(0) = 1 y(0) = 4 4. [5 points] y′ = 9y − y3 y(0) = −4 y(0) = −1 y(0) = 2 y(0) = 4

Sample Paper For Above instruction

This paper aims to analyze and graph solutions of specified autonomous differential equations utilizing the qualitative theory of differential equations. The focus will be on understanding the behavior of solutions, identifying equilibrium points, and sketching qualitative graph features including yz-graphs and ty-graphs for each differential equation with given initial conditions.

Introduction

Autonomous differential equations are fundamental in modeling systems where the rate of change depends solely on the current state, not explicitly on time. The qualitative analysis provides insight into the stability and long-term behavior of solutions without requiring explicit solutions. In this paper, we examine four differential equations each with specified initial conditions and analyze their solutions graphically and qualitatively.

Analysis of Differential Equation 1

The first differential equation given is y' = 23y - 3. To analyze its behavior, we first identify the equilibrium solutions by setting y' = 0:

0 = 23y - 3  =>  y = 3/23 ≈ 0.1304

This constant solution at y ≈ 0.1304 is an equilibrium point where the solution remains constant if initialized there. To examine stability, consider the sign of y' relative to this equilibrium:

  • For y > 3/23, y' > 0, so the solution increases.
  • For y

Given initial conditions y(0)=3 and y(0)=6: since both exceed 3/23, solutions will increase or decrease towards the equilibrium, depending on initial y-values and the sign of the slope. The qualitative graph thus shows solutions moving towards the equilibrium from above or below, indicating a stable equilibrium typically.

The yz-graph will display y vs. time, with the solution approaching y ≈ 0.1304 over time, starting from initial values 3 and 6.

Analysis of Differential Equation 2

The second differential equation is y' = (y - 1)(y - 5). Setting y' = 0 yields equilibrium solutions:

y = 1 and y = 5

Initial conditions include y(0) = -2, 2, 4, 6. The phase line analysis:

  • When y 0 (since both (y-1) and (y-5) are negative, their product is positive), solutions move upward towards y=1.
  • Between 1 and 5, y'
  • For y > 5, y' > 0, solutions increase, moving away from y=5, indicating y=5 is unstable.

Graphically, solutions with initial y= -2 tend to increase to y=1, whereas y=2 and y=4 tend to decrease toward y=1, and y=6 solutions increase unbounded or move away from y=5, showing the stability of equilibria at y=1 and a potential unstable equilibrium at y=5.

Analysis of Differential Equation 3

Equation: y' = - y² + 10 y - 21, which can be factored or solved for equilibrium points:

- y² + 10 y - 21 = 0  =>  y² - 10 y + 21 = 0  =>  y = [10 ± √(100 - 84)]/2 = [10 ± √16]/2
y = (10 ± 4)/2  =>  y=7 and y=3

These are the equilibrium points. For initial conditions y(0)=1 and y(0)=4:

  • At y=3, the slope y' = 0, indicating an equilibrium.
  • For y 0, solutions increase toward y=3.
  • For 3
  • Above y=7, y'

This behavior indicates y=3 is stable and y=7 is semi-stable, with solutions tending to these equilibria depending on initial conditions.

Analysis of Differential Equation 4

Equation: y' = 9 y - y³, which factors as y' = y(9 - y²). Equilibrium points are at y=0 and y=±3:

0, y=3, y=-3

Initial conditions: y= -4, -1, 2, 4.

  • At y=-4, y' = -4(9 - 16) = -4(-7) = 28 > 0, solutions increase toward y=-3, which is a stable equilibrium.
  • At y=-1, y' = -1(9 - 1) = -1(8) = -8
  • At y=2, y' = 2(9 - 4) = 2(5) = 10 > 0, solutions tend to increase toward y=3, which is stable.
  • At y=4, y' = 4(9 - 16) = 4(-7) = -28

The solutions tend toward the stable equilibria at y=0 or y=3 depending on the initial position, with unstable equilibria at y=±3.

Conclusion

Using the qualitative analysis, we effectively visualize the behavior of solutions for each autonomous differential equation without explicit solutions. Equilibrium points serve as centers of attraction or repulsion, indicating stability properties. The phase-line analysis complements the yz-graphs, illustrating solution trajectories over time, and enhances understanding of the long-term behavior of dynamic systems modeled by these equations.

References

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