Let's solve problem 18 step-by-step as requested. ### Given Differential Equation: \[ y' = (y + 2)(y - 1) \] --- ## **Step 1: Find all equilibrium solutions (part a)** **Equilibrium solutions** occur where \( y' = 0 \). Set \( (y + 2)(y - 1) = 0 \): - \( y + 2 = 0 \implies y = -2 \) - \( y - 1 = 0 \implies y = 1 \) **Explanation:** Equilibrium solutions are the constant solutions where the derivative \( y' \) is zero. Here, that happens when \( y = -2 \) or \( y = 1 \). --- ## **Step 2: Determine where solutions are increasing or decreasing (part b)** Recall: - If \( y' > 0 \), solutions are increasing. - If \( y' < 0 \), solutions are decreasing. Let's analyze the sign of \( (y + 2)(y - 1) \) in the intervals determined by the equilibrium points: 1. **For \( y < -2 \):** - \( y + 2 < 0 \) - \( y - 1 < 0 \) - Product: \( (-)(-) = + \), so **increasing**. 2. **For \( -2 < y < 1 \):** - \( y + 2 > 0 \) - \( y - 1 < 0 \) - Product: \( (+)(-) = - \), so **decreasing**. 3. **For \( y > 1 \):** - \( y + 2 > 0 \) - \( y - 1 > 0 \) - Product: \( (+)(+) = + \), so **increasing**. **Explanation:** We checked the sign of \( y' \) in each interval divided by the equilibrium solutions. This tells us where the solution curves move up or down. --- ## **Final Answers** **(a) The equilibrium solutions are:** \[ y = -2 \quad \text{and} \quad y = 1 \] **(b) The solution curves are:** - **Increasing** for \( y < -2 \) and \( y > 1 \) - **Decreasing** for \( -2 < y < 1 \)

Let's solve problem 18 step-by-step as requested. --- ### Given Differential Equation: \[ y' = (y + 2)(y - 1) \] --- ### **Step 1: Find all equilibrium solutions (part a)** **Explanation:** Equilibrium solutions are constant solutions where the derivative \( y' \) equals zero. So, set the right-hand side equal to zero and solve for \( y \): \[ (y + 2)(y - 1) = 0 \] Which gives: \[ \boxed{ y + 2 = 0 \quad \Rightarrow \quad y = -2 } \] and \[ \boxed{ y - 1 = 0 \quad \Rightarrow \quad y = 1 } \] **Answer (a):** \[ \boxed{ \text{Equilibrium solutions are } \quad y = -2 \quad \text{and} \quad y = 1 } \] --- ### **Step 2: Determine where solutions are increasing or decreasing (part b)** **Explanation:** Solutions are increasing where \( y' > 0 \), decreasing where \( y' < 0 \). The sign of \( y' = (y + 2)(y - 1) \) depends on the factors: - For \( y < -2 \): - \( y + 2 < 0 \) - \( y - 1 < 0 \) - Product: \( (-) \times (-) = (+) \) - For \( -2 < y < 1 \): - \( y + 2 > 0 \) - \( y - 1 < 0 \) - Product: \( (+) \times (-) = (-) \) - For \( y > 1 \): - \( y + 2 > 0 \) - \( y - 1 > 0 \) - Product: \( (+) \times (+) = (+) \) **Answer (b):** - **Increasing** in: - \( y < -2 \) - \( y > 1 \) - **Decreasing** in: - \( -2 < y < 1 \) --- ### **Final Summary:** | Equilibrium solutions (a): | \( y = -2 \), \( y = 1 \) | |------------------------------|-------------------------| | Increasing (b): | \( y < -2 \) and \( y > 1 \) | | Decreasing (b): | \( -2 < y < 1 \) | --- **Note:** The problem also asks about concavity and sketching solutions, but since only parts (a) and (b) are explicitly requested here, this completes the answer for these parts. --- **Final answer:** **(a)** \( y = -2, 1 \) **(b)** Increasing when \( y < -2 \) or \( y > 1 \); decreasing when \( -2 < y < 1 \)

Let's analyze problem 18 step-by-step. --- ### Given Differential Equation: \[ y' = (y + 2)(y - 1) \] --- ### **Step 1: Find all equilibrium solutions (part a)** **Explanation:** Equilibrium solutions occur where \( y' = 0 \). Set the right side equal to zero and solve for \( y \): \[ (y + 2)(y - 1) = 0 \] This yields: \[ \boxed{ y = -2 \quad \text{or} \quad y = 1 } \] **Answer (a):** \[ \boxed{ \text{Equilibrium solutions are } y = -2 \quad \text{and} \quad y = 1 } \] --- ### **Step 2: Determine where solutions are increasing or decreasing (part b)** **Explanation:** - Solutions are increasing where \( y' > 0 \). - Solutions are decreasing where \( y' < 0 \). Analyze the sign of \( y' = (y + 2)(y - 1) \) over the real line, considering the critical points at \( y = -2 \) and \( y = 1 \): - For \( y < -2 \): - \( y + 2 < 0 \) - \( y - 1 < 0 \) - Product: \( (-) \times (-) = (+) \) → **solutions increase** - For \( -2 < y < 1 \): - \( y + 2 > 0 \) - \( y - 1 < 0 \) - Product: \( (+) \times (-) = (-) \) → **solutions decrease** - For \( y > 1 \): - \( y + 2 > 0 \) - \( y - 1 > 0 \) - Product: \( (+) \times (+) = (+) \) → **solutions increase** **Answer (b):** - **Increasing** in: - \( y < -2 \) - \( y > 1 \) - **Decreasing** in: - \( -2 < y < 1 \) --- ### **Final summary:** | Part (a): Equilibrium solutions | \( y = -2, 1 \) | |----------------------------------|-----------------| | Part (b): Increasing solutions | \( y < -2 \) and \( y > 1 \) | | Part (b): Decreasing solutions | \( -2 < y < 1 \) | --- **Note:** Further steps involve concavity and sketching, but for parts (a) and (b), this completes the analysis. **Final answer:** **(a)** \( y = -2, 1 \) **(b)** Increasing when \( y < -2 \) or \( y > 1 \); decreasing when \( -2 < y < 1 \)