Find equillibria of the differential equation. Classify them as sinks or sources. Then represent the VERTICAL phase line using y=whatever to label the pts, and between each point use Ʌ or V and | to create arrows between the points. Remeber, this line is vertical so everything needs to be on a new line. Label every point on the line, belling 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 differential equation:

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

### Step 1: Find the Equilibria

Equilibria occur when \( y' = 0 \), meaning we need to solve:

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

#### Factor the quadratic:
The quadratic \( y^2 + y - 2 \) factors as:

\[
y^2 + y - 2 = (y - 1)(y + 2)
\]

Thus, the equation becomes:

\[
(y - 3)(y - 1)(y + 2) = 0
\]

The equilibria points are the roots of this equation:

\[
y = 3, \quad y = 1, \quad y = -2
\]

### Step 2: Classify the Equilibria (Sinks, Sources, or Nodes)

To classify the equilibria, we need to analyze the sign of \( y' \) in the intervals between the equilibria points.

The critical points divide the \( y \)-axis into four intervals:

1. \( (-\infty, -2) \)
2. \( (-2, 1) \)
3. \( (1, 3) \)
4. \( (3, \infty) \)

We will pick a test point in each interval and check the sign of \( y' \).

- **Interval \( (-\infty, -2) \): Test \( y = -3 \)**  
  \[
  y' = (-3 - 3)(-3 - 1)(-3 + 2) = (-6)(-4)(-1) = -24
  \]
  Since \( y' < 0 \), the solution is decreasing in this interval.

- **Interval \( (-2, 1) \): Test \( y = 0 \)**  
  \[
  y' = (0 - 3)(0 - 1)(0 + 2) = (-3)(-1)(2) = 6
  \]
  Since \( y' > 0 \), the solution is increasing in this interval.

- **Interval \( (1, 3) \): Test \( y = 2 \)**  
  \[
  y' = (2 - 3)(2 - 1)(2 + 2) = (-1)(1)(4) = -4
  \]
  Since \( y' < 0 \), the solution is decreasing in this interval.

- **Interval \( (3, \infty) \): Test \( y = 4 \)**  
  \[
  y' = (4 - 3)(4 - 1)(4 + 2) = (1)(3)(6) = 18
  \]
  Since \( y' > 0 \), the solution is increasing in this interval.

### Equilibrium Classification:

- \( y = -2 \): **Source** (since the solution moves away from it, \( \uparrow \downarrow \))
- \( y = 1 \): **Sink** (since the solution moves toward it, \( \downarrow \uparrow \))
- \( y = 3 \): **Source** (since the solution moves away from it, \( \uparrow \downarrow \))

### Step 3: Vertical Phase Line Representation

We represent the phase line with equilibria at \( y = -2 \), \( y = 1 \), and \( y = 3 \), using arrows to indicate the direction of the solution.

Here’s the vertical phase line:

\[
\begin{aligned}
 & \quad \uparrow \\
 & y = 3 \ (\text{Source}) \\
 & \quad \downarrow \\
 & \uparrow \\
 & y = 1 \ (\text{Sink}) \\
 & \downarrow \\
 & \uparrow \\
 & y = -2 \ (\text{Source}) \\
 & \downarrow \\
\end{aligned}
\]

### Summary of Results:

- \( y = 3 \): Source (solution moves away from this point)
- \( y = 1 \): Sink (solution moves toward this point)
- \( y = -2 \): Source (solution moves away from this point)

---

Would you like to dive into any particular detail of this? Here are 5 related questions:

1. How do you determine if a point is a source or sink mathematically?
2. What happens at an equilibrium point that is neither a source nor a sink?
3. Can a differential equation have more than one sink or source?
4. How does the behavior of the solution change near a saddle point compared to a sink or source?
5. Can you represent this phase line for a more complex system of equations?

**Tip:** To classify equilibria, checking the sign of the derivative \( y' \) in intervals around the equilibrium is crucial for determining the behavior (sink or source).

Find equillibria of the differential equation. Classify them as sinks or sources. Then represent the VERTICAL phase line using y=whatever to label the pts, and between each point use Ʌ or V and | to create arrows between the points. Remeber, this line is vertical so everything needs to be on a new line. Label every point on the line, belling 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)

This detailed guide explores how to find the equilibria of the differential equation y' = (y - 3)(y^2 + y - 2), classify them as sinks or sources, and represent the vertical phase line. Learn the step-by-step process of solving the equation and analyzing the stability of each equilibrium point.

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