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The given differential equation is:

\[
x' = (x + 2.5)(x + 0.5)^3(x - 0.5)^2(x - 2.5)
\]

To find the constant (equilibrium) solutions, we set \( x' = 0 \). This gives:

\[
(x + 2.5)(x + 0.5)^3(x - 0.5)^2(x - 2.5) = 0
\]

The equilibrium solutions are the roots of this equation, i.e., where any of the factors is zero.

### Step 1: Solve for equilibrium points
The factors are:
1. \( x + 2.5 = 0 \) gives \( x = -2.5 \)
2. \( x + 0.5 = 0 \) gives \( x = -0.5 \)
3. \( x - 0.5 = 0 \) gives \( x = 0.5 \)
4. \( x - 2.5 = 0 \) gives \( x = 2.5 \)

Thus, the equilibrium points are:
\[
x = -2.5, -0.5, 0.5, 2.5
\]

### Step 2: Determine stability
To determine whether these equilibrium points are stable, semi-stable, or unstable, we look at the behavior of \( x' \) near these points. We examine the signs of the factors of \( x' \) on intervals around each equilibrium point:

1. **At \( x = -2.5 \)**:
   - The factor \( (x + 2.5) \) changes sign as \( x \) passes through \( -2.5 \). This indicates **stability**.
   
2. **At \( x = -0.5 \)**:
   - The factor \( (x + 0.5)^3 \) does not change sign because it's raised to an odd power, but it behaves like a cubic function, which indicates **semi-stability** (as the derivative is very slow around \( x = -0.5 \)).
   
3. **At \( x = 0.5 \)**:
   - The factor \( (x - 0.5)^2 \) is raised to an even power, so the sign of the product does not change. This indicates **stability**.

4. **At \( x = 2.5 \)**:
   - The factor \( (x - 2.5) \) changes sign, indicating **instability**.

### Final Answer
The equilibrium solutions in increasing order are:
\[
x = -2.5 \ (\text{stable}), \quad x = -0.5 \ (\text{semi-stable}), \quad x = 0.5 \ (\text{stable}), \quad x = 2.5 \ (\text{unstable})
\]

Would you like further clarification or more details on any specific part?

Here are some related questions to expand on:
1. How do we determine stability from the derivative sign changes?
2. What are the physical interpretations of stability in differential equations?
3. How do the powers of the terms influence the stability classification?
4. Can this method be generalized to higher-order differential equations?
5. What happens if there are complex roots in such an equation?

**Tip:** Stability is linked to whether the system returns to equilibrium after a small disturbance. A stable system does, while an unstable one moves away.

Consider the differential equation x' = (x + 2.5)(x + 0.5)^3(x - 0.5)^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.

Discover how to solve the differential equation x' = (x + 2.5)(x + 0.5)^3(x - 0.5)^2(x - 2.5) by finding the equilibrium solutions and analyzing their stability. This solution involves understanding equilibrium points and their stability based on sign changes in the derivative. Suitable for calculus and differential equations students.

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