Consider the differential equation: x'=-(x+3)(x+1)^3(x-1)^2(x-2.5). List the constant (or equilibrium) solutions to this differential equation in increasing order and indicate whether or not these equilibria are stable, semi-stable, or unstable.

The differential equation given is:

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

To find the **equilibrium solutions**, we need to solve \(x' = 0\). This occurs when:

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

The solutions to this equation are the values of \(x\) where any factor on the right-hand side equals zero. These solutions (equilibria) are:

\[
x = -3, \quad x = -1, \quad x = 1, \quad x = 2.5
\]

### Classifying the Stability of Equilibria

The stability of each equilibrium can be determined by analyzing the behavior of \(x'\) around each equilibrium point:

- **Stable equilibrium**: Solutions tend to move toward the equilibrium from both sides.
- **Unstable equilibrium**: Solutions tend to move away from the equilibrium.
- **Semi-stable equilibrium**: Solutions move toward the equilibrium from one side and away from the other side.

We will examine the multiplicity of each root and analyze the sign of \(x'\) near these points.

1. **At \(x = -3\):**
   - The factor \((x + 3)\) appears with an exponent of 1 (odd multiplicity).
   - Since the factor \((x + 3)\) crosses the x-axis, the equilibrium at \(x = -3\) is **unstable**.

2. **At \(x = -1\):**
   - The factor \((x + 1)^3\) has an exponent of 3 (odd multiplicity).
   - This odd power suggests that the equilibrium at \(x = -1\) is **unstable**.

3. **At \(x = 1\):**
   - The factor \((x - 1)^2\) has an exponent of 2 (even multiplicity).
   - Since the power is even, the function \(x'\) does not change sign around \(x = 1\). This makes the equilibrium at \(x = 1\) **stable**.

4. **At \(x = 2.5\):**
   - The factor \((x - 2.5)\) has an exponent of 1 (odd multiplicity).
   - This suggests that the equilibrium at \(x = 2.5\) is **unstable**.

### Summary

The equilibrium solutions and their stability classifications are:

- \(x = -3\): Unstable
- \(x = -1\): Unstable
- \(x = 1\): Stable
- \(x = 2.5\): Unstable

Let me know if you'd like more details or have any questions.

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**Related questions:**
1. What are the implications of having unstable equilibria in a physical system?
2. How would the system behave if the factor \((x - 1)^2\) had an odd power?
3. Can you describe the role of multiplicity in determining stability of equilibria?
4. What happens to the solution trajectories as \(x \to \pm \infty\)?
5. How would a small perturbation near a semi-stable equilibrium behave?

**Tip:** The multiplicity of roots directly affects the stability of equilibria in differential equations. An even multiplicity indicates stability, while an odd multiplicity indicates instability.

Analyze the equilibrium solutions of the differential equation x' = -(x+3)(x+1)^3(x-1)^2(x-2.5), classify their stability, and understand how root multiplicity influences equilibrium behavior. This solution covers stability analysis and differential equation concepts for undergraduate mathematics students.

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