Math Problem Statement

Consider the differential equation: x'=-(x+3)(x+1)^3(x-1)^2(x-2.5). list the equilibrium constants solutions to this differential equation. indicate whether or not they are stable, unstable or semi-stable.

Solution

The given differential equation is:

x=(x+3)(x+1)3(x1)2(x2.5)x' = -(x + 3)(x + 1)^3(x - 1)^2(x - 2.5)

To find the equilibrium points, we set x=0x' = 0, which occurs when any of the factors on the right-hand side is zero:

(x+3)(x+1)3(x1)2(x2.5)=0(x + 3)(x + 1)^3(x - 1)^2(x - 2.5) = 0

The equilibrium points are the solutions to:

x+3=0or(x+1)3=0or(x1)2=0or(x2.5)=0x + 3 = 0 \quad \text{or} \quad (x + 1)^3 = 0 \quad \text{or} \quad (x - 1)^2 = 0 \quad \text{or} \quad (x - 2.5) = 0

Thus, the equilibrium points are:

  • x=3x = -3
  • x=1x = -1
  • x=1x = 1
  • x=2.5x = 2.5

Stability Analysis

To determine the stability of each equilibrium point, we analyze the sign changes of xx' around these points. We use the multiplicities of the factors (odd or even powers) to infer the stability:

  1. x=3x = -3:

    • The factor (x+3)(x + 3) has a multiplicity of 1, which is odd.
    • Odd multiplicities indicate a change in the sign of xx' at this point.
    • Conclusion: Unstable (because the sign of xx' flips at x=3x = -3).
  2. x=1x = -1:

    • The factor (x+1)3(x + 1)^3 has a multiplicity of 3, which is odd.
    • Odd multiplicities indicate a change in the sign of xx' at this point.
    • Conclusion: Unstable.
  3. x=1x = 1:

    • The factor (x1)2(x - 1)^2 has a multiplicity of 2, which is even.
    • Even multiplicities indicate that xx' does not change sign at this point.
    • Conclusion: Semi-stable (since the solution approaches from one side but does not cross).
  4. x=2.5x = 2.5:

    • The factor (x2.5)(x - 2.5) has a multiplicity of 1, which is odd.
    • Odd multiplicities indicate a change in the sign of xx' at this point.
    • Conclusion: Stable (because xx' flips from negative to positive at this point).

Summary of Equilibrium Points and Their Stability:

  • x=3x = -3: Unstable
  • x=1x = -1: Unstable
  • x=1x = 1: Semi-stable
  • x=2.5x = 2.5: Stable

Would you like further details on the stability analysis? Here are some related questions:

  1. What happens to the solution behavior near unstable equilibrium points?
  2. Can you explain why odd multiplicity corresponds to instability?
  3. How does the solution behave around semi-stable points?
  4. What is the importance of equilibrium stability in real-world systems?
  5. How would the stability change if the multiplicity of a factor was different?

Tip: For stability analysis of equilibrium points, remember that odd powers usually result in sign changes (unstable/stable), while even powers lead to no sign change (semi-stable).

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Equilibrium Points
Stability Analysis

Formulas

Equilibrium condition: Set the derivative x' = 0 to find equilibrium points
Stability analysis based on the sign change of x' around the equilibrium points

Theorems

Stability Theorem: Odd multiplicity of a factor results in a sign change (unstable/stable); even multiplicity leads to no sign change (semi-stable)

Suitable Grade Level

Advanced High School or College (Grades 11-12, Undergraduate)