Let's solve the problem step by step as requested: --- ### **Step 1: Find the equilibrium points** **Equation:** \[ x' = x^2 - 3ax + 2a^2 \] **Equilibrium points** occur where \( x' = 0 \): \[ x^2 - 3ax + 2a^2 = 0 \] This is a quadratic equation in \( x \): \[ x = \frac{3a \pm \sqrt{(3a)^2 - 4 \cdot 1 \cdot 2a^2}}{2} \] Calculate the discriminant: \[ (3a)^2 - 4 \cdot 1 \cdot 2a^2 = 9a^2 - 8a^2 = a^2 \] So, \[ x = \frac{3a \pm |a|}{2} \] This gives two solutions: - \( x_1 = \frac{3a + |a|}{2} \) - \( x_2 = \frac{3a - |a|}{2} \) **Explanation:** The equilibrium points are found by setting the right-hand side to zero and solving the resulting quadratic equation for \( x \). --- ### **Step 2: Sketch phase portraits for different values of \( a \)** #### **Case 1: \( a > 0 \)** - \( |a| = a \), so: - \( x_1 = \frac{3a + a}{2} = 2a \) - \( x_2 = \frac{3a - a}{2} = a \) - So, equilibria: \( x = a \), \( x = 2a \) #### **Case 2: \( a = 0 \)** - Both equilibria collapse to zero: - \( x_1 = x_2 = 0 \) #### **Case 3: \( a < 0 \)** - \( |a| = -a \), so: - \( x_1 = \frac{3a + (-a)}{2} = \frac{2a}{2} = a \) - \( x_2 = \frac{3a - (-a)}{2} = \frac{4a}{2} = 2a \) - So, equilibria: \( x = a \), \( x = 2a \) (but both negative) **Phase Portraits:** - The sign of \( x' \) between and outside the equilibria determines the direction of flow. - The equilibrium points will switch stability depending on the sign of \( a \). **Summary Table:** | \( a \) | Equilibria | Relative position | |----------------|-------------------|------------------------| | \( a > 0 \) | \( x = a, 2a \) | \( a < 2a \) | | \( a = 0 \) | \( x = 0 \) | (double root) | | \( a < 0 \) | \( x = 2a, a \) | \( 2a < a \) | **Explanation:** Depending on the value of \( a \), the nature and stability of equilibrium points change, and so do the phase portraits. For \( a > 0 \), the equilibria are both positive; for \( a < 0 \), they're both negative; for \( a = 0 \), there's a single degenerate equilibrium at 0. --- ## **Final Answer** ### (a) **The equilibrium points are:** \[ x = \frac{3a \pm |a|}{2} \] ### (b) **Phase portraits:** - For \( a > 0 \): Two equilibria at \( x = a \) and \( x = 2a \) (both positive). - For \( a = 0 \): One double equilibrium at \( x = 0 \). - For \( a < 0 \): Two equilibria at \( x = 2a \) and \( x = a \) (both negative). The stability and direction of flow between these points can be deduced by analyzing the sign of \( x' \) in each region.