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.