Let's break down the question step by step: ### Given: **Differential Equation:** \[ \frac{dy}{dx} = \cos y \] **Interval:** \[ -3 \leq x \leq 3 \] **Points for solution curve:** \[ (0, \pi/2),\, (-1, 0) \] --- ### **Step 1: Construct the Slope Field** The slope field is a graphical representation showing the slope (direction) at each point \((x, y)\). Here, the slope at each point is given by \(\cos y\). - For each point \((x, y)\), calculate the slope \(m = \cos y\). - The slope only depends on \(y\), **not on \(x\)**. #### **Key Slope Values:** - When \(y = 0\), \(\cos(0) = 1\) (slope = 1, rising line). - When \(y = \pi/2\), \(\cos(\pi/2) = 0\) (slope = 0, horizontal line). - When \(y = \pi\), \(\cos(\pi) = -1\) (slope = -1, falling line). - When \(y = 3\pi/2\), \(\cos(3\pi/2) = 0\) (slope = 0, horizontal line). So, for each horizontal line at \(y\), the slope is the same for all \(x\). --- ### **Step 2: Sketch the Slope Field** You can sketch this as a grid of short line segments. For each \(y\): - At \(y = 0\): draw lines with slope 1. - At \(y = \pi/2\): draw horizontal lines (slope 0). - At \(y = \pi\): draw lines with slope -1. - Repeat this for several values of \(y\). --- ### **Step 3: Find the Solution Curve** #### **Separate Variables:** \[ \frac{dy}{dx} = \cos y \] \[ \frac{dy}{\cos y} = dx \] #### **Integrate Both Sides:** \[ \int \sec y\, dy = \int dx \] Recall: \[ \int \sec y\, dy = \ln |\sec y + \tan y| \] So, \[ \ln |\sec y + \tan y| = x + C \] #### **Solve for y:** \[ |\sec y + \tan y| = e^{x+C} = Ae^x \] Where \(A = \pm e^C\), a constant. --- ### **Step 4: Apply Initial Conditions** #### **Point (0, \(\frac{\pi}{2}\)):** - \(x = 0, y = \pi/2\) - \(\sec(\pi/2) = \infty\), \(\tan(\pi/2) = \infty\) (undefined!) So this is a **vertical asymptote** in your solution: the solution curve approaches \(y = \pi/2\) but never crosses it. #### **Point (-1, 0):** - \(x = -1, y = 0\) - \(\sec(0) = 1\), \(\tan(0) = 0\) - \(|1+0| = Ae^{-1}\) ⇒ \(A = e^{-1}\) So the solution is: \[ |\sec y + \tan y| = e^{-1}e^x = e^{x-1} \] Or, remove absolute value (since \(\sec y + \tan y > 0\) near \(y = 0\)): \[ \sec y + \tan y = e^{x-1} \] --- ### **Step 5: Sketch the Solution Curve** - For \((0, \pi/2)\): The solution curve will be a vertical line at \(x = 0\) where \(y \to \pi/2\), **but not defined at exactly \(y = \pi/2\)**. - For \((-1, 0)\): Plot the curve \(\sec y + \tan y = e^{x-1}\). #### **How to Sketch:** - Draw the slope field. - At each point along \((-1, 0)\) use the curve above. - The solution starting at \((-1, 0)\) will increase in \(y\) as \(x\) increases, approaching \(y = \pi/2\) as \(x\) approaches a certain value. --- ### **Final Answer (Summary):** **1. The slope field for \(\frac{dy}{dx} = \cos y\) is drawn using the slope at each point \((x, y)\), which depends only on \(y\):** - Slope = 1 at \(y = 0\), - Slope = 0 at \(y = \pi/2, 3\pi/2, ...\), - Slope = -1 at \(y = \pi\), etc. **2. The solution curve through \((-1, 0)\) is:** \[ \sec y + \tan y = e^{x-1} \] **3. The solution curve through \((0, \pi/2)\) is undefined at exactly that point, but curves approach \(y = \pi/2\) as \(x\) increases.** --- **If you need a hand-drawn or plotted slope field, let me know and I can generate one!**