Let's break this down into two steps as requested. --- ### Step 1: **Determine the Regions of Concavity** To determine where the solution curves are concave up or concave down, **we need the differential equation** that these solutions satisfy (it is not provided in your image). Typically, for a differential equation of the form \(\frac{dy}{dx} = f(x, y)\): - The concavity of solution curves is determined by the sign of the second derivative \(\frac{d^2y}{dx^2}\). - Compute: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} \] - The curve is **concave up** where \(\frac{d^2y}{dx^2} > 0\) and **concave down** where \(\frac{d^2y}{dx^2} < 0\). **Explanation:** Without the explicit form of \(f(x, y)\), we can't find the exact regions, but the method is: 1. Find \(\frac{d^2y}{dx^2}\). 2. Set \(\frac{d^2y}{dx^2} > 0\) for concave up, and \(\frac{d^2y}{dx^2} < 0\) for concave down. 3. Solve these inequalities to determine the regions in the \(xy\)-plane. --- ### Step 2: **Sketch Representative Solution Curves** - In each region identified in Step 1, sketch solution curves so that: - In regions where the curves are **concave up**, the curves bend upwards (like a cup). - In regions where the curves are **concave down**, the curves bend downwards (like a cap). **Explanation:** The sketch should clearly show the change in concavity at the boundaries between regions, so the solution curves transition smoothly from concave up to concave down or vice versa. --- ### **Final Answer (Summary):** 1. **Find** where the second derivative \(\frac{d^2y}{dx^2}\) is positive (concave up) and negative (concave down) by analyzing the differential equation. 2. **Sketch** solution curves in the \(xy\)-plane, showing upward bending in concave-up regions and downward bending in concave-down regions. **Note:** To give specific regions and an actual sketch, please provide the differential equation! The steps above are the general procedure.

Let's analyze the differential equation \( y' = (y + 2)(y - 1) \) step by step to determine the regions of concavity and sketch representative solutions. --- ### Step 1: Find the second derivative \(\frac{d^2 y}{dx^2}\) to determine concavity. **Explanation:** The concavity of the solution curves depends on the sign of the second derivative \(\frac{d^2 y}{dx^2}\). Since \( y' = f(y) = (y + 2)(y - 1) \), and \(f\) depends only on \( y \), the second derivative is: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( y' \right) = \frac{d}{dx} f(y) = \frac{df}{dy} \cdot y' \] Calculating: \[ \frac{df}{dy} = \frac{d}{dy} \left( (y + 2)(y - 1) \right) \] Expand \(f(y)\): \[ f(y) = y^2 + y - 2 \] Differentiate: \[ \frac{df}{dy} = 2y + 1 \] Now, using the chain rule: \[ \frac{d^2 y}{dx^2} = \frac{df}{dy} \cdot y' = (2y + 1) \cdot (y + 2)(y - 1) \] **Therefore:** \[ \boxed{ \frac{d^2 y}{dx^2} = (2y + 1) \cdot (y + 2)(y - 1) } \] --- ### Step 2: Determine the regions where the solution curves are concave up or down. - The solution is **concave up** where \(\frac{d^2 y}{dx^2} > 0\). - It is **concave down** where \(\frac{d^2 y}{dx^2} < 0\). Identify the zeros of \(\frac{d^2 y}{dx^2}\): \[ (2y + 1) = 0 \implies y = -\frac{1}{2} \] \[ (y + 2) = 0 \implies y = -2 \] \[ (y - 1) = 0 \implies y = 1 \] These values partition the \(y\)-axis into intervals: \[ (-\infty, -2), \quad (-2, -\frac{1}{2}), \quad (-\frac{1}{2}, 1), \quad (1, \infty) \] Test each interval to determine the sign of \(\frac{d^2 y}{dx^2}\): - For \( y < -2 \): pick \( y = -3 \): \[ (2(-3) + 1) = -6 + 1 = -5 < 0 \] \[ (y + 2) = -3 + 2 = -1 < 0 \] \[ (y - 1) = -3 - 1 = -4 < 0 \] Product: \[ (-) \times (-) \times (-) = (-) \times (+) = (-) \implies \text{Negative} \] So, \(\frac{d^2 y}{dx^2} < 0\) → **Concave down** in \((-\infty, -2)\). - For \(-2 < y < -\frac{1}{2}\): pick \( y = -1 \): \[ (2(-1) + 1) = -2 + 1 = -1 < 0 \] \[ (y + 2) = -1 + 2 = 1 > 0 \] \[ (y - 1) = -1 - 1 = -2 < 0 \] Product: \[ (-) \times (+) \times (-) = (+) \implies \text{Positive} \] So, **Concave up** in \((-2, -\frac{1}{2})\). - For \(-\frac{1}{2} < y < 1\): pick \( y = 0 \): \[ (2(0) + 1) = 1 > 0 \] \[ (y + 2) = 0 + 2 = 2 > 0 \] \[ (y - 1) = 0 - 1 = -1 < 0 \] Product: \[ (+) \times (+) \times (-) = (-) \implies \text{Negative} \] So, **Concave down** in \((- \frac{1}{2}, 1)\). - For \( y > 1 \): pick \( y = 2 \): \[ (2(2) + 1) = 4 + 1 = 5 > 0 \] \[ (y + 2) = 2 + 2 = 4 > 0 \] \[ (y - 1) = 2 - 1 = 1 > 0 \] Product: \[ (+) \times (+) \times (+) = (+) \implies \text{Positive} \] So, **Concave up** in \((1, \infty)\). --- ### **Summary of Concavity Regions:** | \( y \)-interval | Concavity | Sign of \(\frac{d^2 y}{dx^2}\) | Region description | |------------------------------|--------------|------------------------------|----------------------------| | \((-\infty, -2)\) | Concave down | Negative | y less than -2 | | \((-2, -\frac{1}{2})\) | Concave up | Positive | between -2 and \(-\frac{1}{2}\) | | \((- \frac{1}{2}, 1)\) | Concave down | Negative | between \(-\frac{1}{2}\) and 1 | | \((1, \infty)\) | Concave up | Positive | y greater than 1 | --- ### **Sketching representative solution curves:** - In the **concave up** regions (\((-2, -\frac{1}{2})\) and \((1, \infty)\)), solution curves bend upward. - In the **concave down** regions (\((-\infty, -2)\) and \((- \frac{1}{2}, 1)\)), they bend downward. - The curves should transition smoothly at the boundary lines \( y = -2, -\frac{1}{2}, 1 \), reflecting the change in concavity. --- **Final note:** The exact shape of the solutions depends on initial conditions, but the concavity analysis indicates where the curves are "cup-shaped" or "cap-shaped."

Let's analyze the differential equation \( y' = (y + 2)(y - 1) \) step by step. --- ### (a) Find all equilibrium solutions **Step 1:** Equilibrium solutions occur where \( y' = 0 \). Set: \[ (y + 2)(y - 1) = 0 \] **Solve:** \[ y + 2 = 0 \implies y = -2 \] \[ y - 1 = 0 \implies y = 1 \] **Answer:** **Equilibrium solutions are the constant solutions \( y = -2 \) and \( y = 1 \).** --- ### (b) Determine regions where solutions are increasing or decreasing **Step 2:** Sign of \( y' = (y + 2)(y - 1) \) determines whether solutions are increasing or decreasing. - **Solutions increase where \( y' > 0 \).** - **Solutions decrease where \( y' < 0 \).** **Step 3:** Analyze the sign of \( y' \) based on \( y \): - Zeros at \( y = -2 \) and \( y = 1 \). These partition the \( y \)-axis into intervals: \[ (-\infty, -2), \quad (-2, 1), \quad (1, \infty) \] - **Test points:** | Interval | Test \( y \) | Sign of \( y' \) | Explanation | |------------|--------------|------------------|------------------------------| | \( (-\infty, -2) \) | \( y = -3 \) | \( (-3+2)(-3-1) = (-1)(-4) = 4 > 0 \) | Increasing in \( y < -2 \) | | \( (-2, 1) \) | \( y = 0 \) | \( (0+2)(0-1) = (2)(-1) = -2 < 0 \) | Decreasing in \( -2 < y < 1 \) | | \( (1, \infty) \) | \( y = 2 \) | \( (2+2)(2-1) = (4)(1) = 4 > 0 \) | Increasing in \( y > 1 \) | --- ### **Answer:** - **Solutions are increasing in** \(\boxed{(-\infty, -2)} \text{ and } (1, \infty)\). - **Solutions are decreasing in** \(\boxed{(-2, 1)}\). --- ### (c) Determine regions where solution curves are concave up or down **Step 4:** The concavity depends on the second derivative \(\frac{d^2 y}{dx^2}\). Since \( y' = f(y) = (y+2)(y-1) \), and \(f\) depends only on \( y \), the second derivative is: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} y' = \frac{d}{dy}f(y) \cdot y' = f'(y) \cdot y' \] Calculate: \[ f'(y) = \frac{d}{dy} [(y+2)(y-1)] = \frac{d}{dy} (y^2 + y - 2) = 2y + 1 \] Thus: \[ \boxed{ \frac{d^2 y}{dx^2} = (2y + 1) \cdot (y+2)(y-1) } \] **Step 5:** Find the zeros of \(\frac{d^2 y}{dx^2}\): \[ 2y + 1 = 0 \implies y = -\frac{1}{2} \] \[ (y+2) = 0 \implies y = -2 \] \[ (y-1) = 0 \implies y = 1 \] These points partition the \( y \)-axis: \[ (-\infty, -2), \quad (-2, -\frac{1}{2}), \quad (-\frac{1}{2}, 1), \quad (1, \infty) \] **Step 6:** Sign of \(\frac{d^2 y}{dx^2}\): - **Test points:** | Interval | \( y \) | \( 2y+1 \) | \( y+2 \) | \( y-1 \) | Product sign | Concavity | |------------|---------|------------|-----------|----------|--------------|------------| | \((-\infty, -2)\) | \( y=-3 \) | \( 2(-3)+1=-6+1=-5 \) | \(-3+2=-1\) | \(-3-1=-4\) | \((-)\times(-)\times(-)= - \) | Concave down | | \((-2, -\frac{1}{2})\) | \( y=-1 \) | \( -2+1=-1 \) | \( 1 \) | \(-1-1=-2\) | \( (-)\times(+) \times(-)= + \) | Concave up | | \((- \frac{1}{2}, 1)\) | \( y=0 \) | \( 1 \) | \( 2 \) | \(-1\) | \( (+)\times(+)\times(-)= - \) | Concave down | | \( (1, \infty) \) | \( y=2 \) | \( 4+1=5 \) | \( 4 \) | \( 2-1=1 \) | \( (+)\times(+)\times(+)= + \) | Concave up | --- ### **Answer:** | \( y \)-interval | Concavity | Sign of \(\frac{d^2 y}{dx^2}\) | Region description | |------------------------------|--------------|------------------------------|----------------------------| | \((-\infty, -2)\) | Concave down | Negative | y less than -2 | | \((-2, -\frac{1}{2})\) | Concave up | Positive | between -2 and \(-\frac{1}{2}\) | | \((- \frac{1}{2}, 1)\) | Concave down | Negative | between \(-\frac{1}{2}\) and 1 | | \((1, \infty)\) | Concave up | Positive | y greater than 1 | --- ### (d) Sketch representative solution curves - **In regions where solutions are increasing:** - For \( y < -2 \), the solutions increase and are **concave down**. - For \( y > 1 \), solutions increase and are **concave up**. - **In regions where solutions are decreasing:** - For \( -2 < y < 1 \), solutions decrease; the concavity switches at \( y = -\frac{1}{2} \). - Curves bend **downward** in \((-2, -\frac{1}{2})\) (concave up) and **upward** in \((- \frac{1}{2}, 1)\) (concave down). - **At the equilibrium lines \( y=-2 \) and \( y=1 \),** solutions can approach these lines asymptotically. - **Summary of shape:** - In \( y < -2 \): solutions increase and bend downward. - In \( -2 < y < -\frac{1}{2} \): solutions decrease and bend upward. - In \( -\frac{1}{2} < y < 1 \): solutions decrease and bend downward. - In \( y > 1 \): solutions increase and bend upward. --- **Final note:** To sketch specific solutions, pick initial points in each region and plot the curves respecting the increasing/decreasing and concavity behaviors described. The solutions will tend toward the equilibrium solutions \( y = -2 \) or \( y=1 \) depending on initial conditions.