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.

The given differential equation is:

\[
x' = -(x + 3)(x + 1)^3(x - 1)^2(x - 2.5)
\]

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

\[
(x + 3)(x + 1)^3(x - 1)^2(x - 2.5) = 0
\]

The equilibrium points are the solutions to:

\[
x + 3 = 0 \quad \text{or} \quad (x + 1)^3 = 0 \quad \text{or} \quad (x - 1)^2 = 0 \quad \text{or} \quad (x - 2.5) = 0
\]

Thus, the equilibrium points are:

- \(x = -3\)
- \(x = -1\)
- \(x = 1\)
- \(x = 2.5\)

### Stability Analysis

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

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

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

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

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

### Summary of Equilibrium Points and Their Stability:
- \(x = -3\): **Unstable**
- \(x = -1\): **Unstable**
- \(x = 1\): **Semi-stable**
- \(x = 2.5\): **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).

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

Analyze the equilibrium points of the differential equation x' = -(x + 3)(x + 1)^3(x - 1)^2(x - 2.5) and determine their stability. Learn how odd and even multiplicities influence stability, with detailed explanations suitable for advanced high school and undergraduate students.

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