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).