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Let's analyze the differential equation given:

\[
\frac{dy}{dx} = y(y^2 - 4)
\]

The equation has three equilibrium points at \( y = 0 \), \( y = -2 \), and \( y = 2 \). To find the stability of each equilibrium, we need to examine the behavior of the derivative \( \frac{dy}{dx} \) around these points.

### Step 1: Factor the equation
\[
\frac{dy}{dx} = y(y - 2)(y + 2)
\]

### Step 2: Determine the sign of \( \frac{dy}{dx} \) in different regions
We check the sign of \( \frac{dy}{dx} \) in the intervals around the equilibrium points \( y = -2 \), \( y = 0 \), and \( y = 2 \).

- **For \( y < -2 \):**  
  All terms \( y \), \( y-2 \), and \( y+2 \) are negative, so \( \frac{dy}{dx} > 0 \). This means the solution is increasing for \( y < -2 \).

- **For \( -2 < y < 0 \):**  
  Here, \( y \) is negative, \( y-2 \) is negative, and \( y+2 \) is positive, so \( \frac{dy}{dx} < 0 \). This means the solution is decreasing for \( -2 < y < 0 \).

- **For \( 0 < y < 2 \):**  
  Now \( y \) is positive, \( y-2 \) is negative, and \( y+2 \) is positive, so \( \frac{dy}{dx} < 0 \). The solution is decreasing for \( 0 < y < 2 \).

- **For \( y > 2 \):**  
  In this case, \( y \), \( y-2 \), and \( y+2 \) are all positive, so \( \frac{dy}{dx} > 0 \). The solution is increasing for \( y > 2 \).

### Step 3: Stability of equilibrium points
- **At \( y = -2 \):**  
  The derivative changes from positive (increasing) to negative (decreasing), meaning \( y = -2 \) is stable.

- **At \( y = 0 \):**  
  The derivative does not change its sign (decreasing on both sides), meaning \( y = 0 \) is unstable.

- **At \( y = 2 \):**  
  The derivative changes from negative (decreasing) to positive (increasing), meaning \( y = 2 \) is stable.

### Step 4: Correct statement
The correct answer is:
\[
\boxed{y = -2 \text{ is stable and } y = 0 \text{ and } y = 2 \text{ are stable}}
\]

Do you have any questions or need further details on any steps?

Here are 5 related questions to expand on this:

1. How can you apply linear stability analysis to verify stability of equilibrium points?
2. What happens if we modify the differential equation slightly, such as by changing the powers of \( y \)?
3. How would we analyze the stability if the equation was non-autonomous (explicitly time-dependent)?
4. How does phase plane analysis help in visualizing the stability of systems of equations?
5. How do we classify stability when equilibrium points are neither stable nor unstable?

**Tip:** For stability analysis, a simple sketch of the direction of flow on a phase line can often quickly indicate the behavior of the system near equilibrium points.

The differential equation dy/dx = y(y^2 - 4) has three equilibria y = 0, y = -2, and y = 2. Plot dy/dx against y and find the stability of each of these equilibria. Which of these statements is correct?

Understand the stability analysis for the differential equation dy/dx = y(y^2 - 4). Explore how to find the stability of equilibrium points y = 0, y = -2, and y = 2 by plotting dy/dx against y and analyzing the derivative's behavior.

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