Find equillibria of the differential equation. Classify them as sinks or sources. Then sketch the phase line using ASCII characters, labelleling every point on the line, labelleing all sinks or sources, and adding clear arrows ON THE LINE. Mattter of fact, the line should be MADE of arrows. It shouldn't even be completely connected. Just 2 arrows between each point

y′=(y−3)(y^2+y−2)

We are given the first-order differential equation:

\[
y' = (y - 3)(y^2 + y - 2)
\]

### Step 1: Find the Equilibria
To find the equilibrium points, set \( y' = 0 \). This gives us:

\[
(y - 3)(y^2 + y - 2) = 0
\]

We now solve for \( y \) in each factor.

1. **First factor: \( y - 3 = 0 \)**
   - Solution: \( y = 3 \)

2. **Second factor: \( y^2 + y - 2 = 0 \)**
   - Solve using the quadratic formula:
   \[
   y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}
   \]
   - The roots are:
     \[
     y = \frac{-1 + 3}{2} = 1 \quad \text{and} \quad y = \frac{-1 - 3}{2} = -2
     \]

Thus, the equilibria points are \( y = 3 \), \( y = 1 \), and \( y = -2 \).

### Step 2: Classify the Equilibria

To classify each equilibrium point as a sink (stable), source (unstable), or node (semi-stable), we analyze the sign of \( y' \) in each region divided by these equilibrium points: \( y < -2 \), \( -2 < y < 1 \), \( 1 < y < 3 \), and \( y > 3 \).

#### Analyze the sign of \( y' \):
1. **For \( y < -2 \)**: Choose \( y = -3 \), then:
   \[
   y' = (-3 - 3)((-3)^2 + (-3) - 2) = (-6)(9 - 3 - 2) = (-6)(4) = -24
   \]
   - Therefore, \( y' < 0 \), meaning \( y \) is decreasing.

2. **For \( -2 < y < 1 \)**: Choose \( y = 0 \), then:
   \[
   y' = (0 - 3)((0)^2 + 0 - 2) = (-3)(-2) = 6
   \]
   - Therefore, \( y' > 0 \), meaning \( y \) is increasing.

3. **For \( 1 < y < 3 \)**: Choose \( y = 2 \), then:
   \[
   y' = (2 - 3)((2)^2 + 2 - 2) = (-1)(4 + 2 - 2) = (-1)(4) = -4
   \]
   - Therefore, \( y' < 0 \), meaning \( y \) is decreasing.

4. **For \( y > 3 \)**: Choose \( y = 4 \), then:
   \[
   y' = (4 - 3)((4)^2 + 4 - 2) = (1)(16 + 4 - 2) = (1)(18) = 18
   \]
   - Therefore, \( y' > 0 \), meaning \( y \) is increasing.

#### Classification:
- At \( y = -2 \), the solution is decreasing to the left and increasing to the right, so this is a **sink** (stable).
- At \( y = 1 \), the solution is increasing to the left and decreasing to the right, so this is a **source** (unstable).
- At \( y = 3 \), the solution is decreasing to the left and increasing to the right, so this is a **source** (unstable).

### Step 3: Sketch the Phase Line

Now we can represent the phase line. Using ASCII characters, we'll make the line out of arrows, with 2 arrows between each point:

```
y = -2   y = 1   y = 3
SINK      SOURCE   SOURCE
←←      →→     ←←      →→
```


- \( y = -2 \) is a sink, where arrows point inward.
- \( y = 1 \) and \( y = 3 \) are sources, where arrows point outward.

### Summary:
- **Equilibria**: \( y = -2 \) (sink), \( y = 1 \) (source), \( y = 3 \) (source)
- **Phase Line**: Depicted above.

Would you like more details or clarification on any part of this?

---
Here are 5 related questions:
1. How does linear stability analysis work for classifying equilibria?
2. Can you explain why sources and sinks behave differently in dynamical systems?
3. How do phase lines help in visualizing differential equation solutions?
4. What changes if the differential equation has nonlinear terms of higher degree?
5. Can we extend this analysis to systems of differential equations?

**Tip**: To analyze stability, you can also take the derivative of \( y' \) with respect to \( y \) and check its sign near the equilibria.

Find equilibria of the differential equation. Classify them as sinks or sources. Then sketch the phase line using ASCII characters, labelling every point on the line, labelling all sinks or sources, and adding clear arrows ON THE LINE. Matter of fact, the line should be MADE of arrows. It shouldn't even be completely connected. Just 2 arrows between each point.

y′ = (y−3)(y^2 + y−2)

Explore how to find and classify the equilibrium points of the differential equation y' = (y−3)(y^2 + y−2). This solution includes a stability analysis of the equilibria, identifying sinks and sources, and presents a phase line sketch using ASCII characters. Suitable for college-level calculus students.

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