6. Consider the differential equation \[ \frac{d y}{d t}=y(y-1)^{2}(y-2)^{3}(y-3)^{4} \] Select the equilibrium point(s) in the list below that are nodes: \[ y=0, \quad y=1, \quad y=2, \quad y=3 \]

In the given differential equation: \[ \frac{d y}{d t} = y(y-1)^{2}(y-2)^{3}(y-3)^{4} \] the equilibrium points occur where \(\frac{dy}{dt} = 0\), which are \(y = 0\), \(y = 1\), \(y = 2\), and \(y = 3\). ### Understanding the Nature of Each Equilibrium Point 1. **\(y = 0\) (Multiplicity 1):** - **Behavior:** - For \(y < 0\), \(\frac{dy}{dt} > 0\). - For \(y > 0\), \(\frac{dy}{dt} > 0\). - **Interpretation:** The solution does not cross \(y = 0\) and flows away from it on both sides, indicating it's a **repelling equilibrium**. 2. **\(y = 1\) (Multiplicity 2):** - **Behavior:** - For \(y < 1\), \(\frac{dy}{dt} > 0\). - For \(y > 1\), \(\frac{dy}{dt} > 0\). - **Interpretation:** The solution does not cross \(y = 1\) and flows away on both sides, making it a **non-crossing equilibrium**. Due to its even multiplicity, it behaves like a **node**. 3. **\(y = 2\) (Multiplicity 3):** - **Behavior:** - For \(y < 2\), \(\frac{dy}{dt} < 0\). - For \(y > 2\), \(\frac{dy}{dt} > 0\). - **Interpretation:** The solution crosses \(y = 2\), changing direction from decreasing to increasing. This indicates a **semi-stable equilibrium**. 4. **\(y = 3\) (Multiplicity 4):** - **Behavior:** - For \(y < 3\), \(\frac{dy}{dt} > 0\). - For \(y > 3\), \(\frac{dy}{dt} > 0\). - **Interpretation:** Similar to \(y = 1\), the solution does not cross \(y = 3\) and flows away on both sides, behaving like a **node** due to its even multiplicity. ### Conclusion **Equilibrium points with even multiplicities (i.e., \(y = 1\) and \(y = 3\)) behave as nodes** in the phase line analysis of this differential equation. **Answer:** Both y = 1 and y = 3 are equilibrium points that are nodes.