Let's tackle each problem **step-by-step**. For each, we'll: 1. **Write the differential equation.** 2. **Describe how to sketch the direction field at the specified resolution.** 3. **Explain how to sketch the particular solution(s) passing through the given points.** 4. **(If possible) Give the analytical solution.** --- ## **5a. \( y' = 1 + y^2 \), Point: (1, 1), Plot limits: \( 0 \leq x \leq 3, -3 \leq y \leq 3 \), Resolution: 0.5** ### **Step 1: Write the equation** \( y' = 1 + y^2 \) ### **Step 2: Direction Field** - At each grid point \((x, y)\) (with spacings of 0.5), calculate the slope \( y' = 1 + y^2 \). - The slope depends **only on \(y\)**, not \(x\). - For example, at \(y = 0\): \(y' = 1\) (a 45-degree line). - At \(y = 1\): \(y' = 2\). - At \(y = -1\): \(y' = 2\). - For large \(|y|\), \(y' \) gets very steep. ### **Step 3: Particular Solution through (1,1)** - Start at point (1,1) and sketch a curve that follows the direction field. - Since \(y' > 0\) everywhere, all solution curves are increasing functions. ### **Step 4: Analytical Solution (Optional)** The equation is separable: \[ \frac{dy}{dx} = 1 + y^2 \implies \int \frac{1}{1 + y^2} dy = \int dx \implies \arctan(y) = x + C \] Using the point (1,1): \[ \arctan(1) = 1 + C \implies \frac{\pi}{4} = 1 + C \implies C = \frac{\pi}{4} - 1 \] So, the particular solution is: \[ \arctan(y) = x + \left(\frac{\pi}{4} - 1\right) \] Or, \[ y = \tan\left(x + \frac{\pi}{4} - 1\right) \] --- ## **5b. \( y' = 1 - y^2 \), Points: (0,0), (0,2), Plot limits: \( 0 \leq x \leq 3, -3 \leq y \leq 3 \), Resolution: 0.5** ### **Step 1: Write the equation** \( y' = 1 - y^2 \) ### **Step 2: Direction Field** - At each grid point \((x, y)\), calculate \( y' = 1 - y^2 \). - For \(y = 0\): \(y' = 1\) (45-degree line). - For \(y = 1\) or \(y = -1\): \(y' = 0\) (horizontal segment). - For \(|y| > 1\): \(y'\) is negative (slopes going down). ### **Step 3: Particular Solutions through (0,0) and (0,2)** - Start at (0,0) and (0,2), sketch curves following the direction field. - At \(y = 1\), the field is horizontal—so there is an equilibrium solution at \(y = 1\) and \(y = -1\). ### **Step 4: Analytical Solution (Optional)** This is also separable: \[ \frac{dy}{dx} = 1 - y^2 \implies \int \frac{1}{1 - y^2} dy = \int dx \] \[ \int \left( \frac{1}{2} \left( \frac{1}{1-y} + \frac{1}{1+y} \right) \right) dy = x + C \] \[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x + C \] Use the points to solve for \(C\): - For (0,0): \( \frac{1}{2} \ln(1) = 0 + C \implies C=0 \) So the solution through (0,0) is \[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x \] Or, \[ y = \frac{e^{2x}-1}{e^{2x}+1} \] --- ## **5c. \( y y' + x = 0 \), Point: (0,1), Plot limits: \( -2 \leq x \leq 2, -2 \leq y \leq 2 \), Resolution: 0.5** ### **Step 1: Write the equation** \( y y' + x = 0 \implies y y' = -x \implies y' = -\frac{x}{y} \) ### **Step 2: Direction Field** - At each grid point \((x, y)\), calculate \( y' = -\frac{x}{y} \) - If \(y = 0\), slope is undefined (vertical). - For \(x = 0\), \(y' = 0\) (horizontal). ### **Step 3: Particular Solution through (0,1)** - Start at (0,1), sketch the curve following the direction field. ### **Step 4: Analytical Solution (Optional)** This is separable: \[ y \frac{dy}{dx} = -x \implies y\,dy = -x\,dx \] \[ \int y\,dy = -\int x\,dx \implies \frac{y^2}{2} = -\frac{x^2}{2} + C \] \[ y^2 + x^2 = 2C \] Using (0,1): \[ (0)^2 + (1)^2 = 2C \implies C = \frac{1}{2} \] So the particular solution: \[ y^2 + x^2 = 1 \] (a circle of radius 1 centered at (0,0)). --- ## **Summary of Steps for Each:** 1. **Set up a grid** (with the specified resolution). 2. **At each grid point**, calculate the slope as per the equation. 3. **Draw a small line segment** at that slope. 4. **From the specified initial point(s)**, sketch the curve that follows the direction field. 5. **(Optional)**: Write the analytical solution if needed. --- If you need **actual hand-drawn sketches** or further explanation on how to plot these by hand, let me know!

**Final Answers:** ### **5a. \( y' = 1 + y^2 \), passing through (1,1)** **Solution:** \(\arctan(y) = x + \frac{\pi}{4} - 1\) **Explicit form:** \( y = \tan\left( x + \frac{\pi}{4} - 1 \right) \) --- ### **5b. \( y' = 1 - y^2 \), passing through (0,0) and (0,2)** **Solution:** \(\frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| = x + C\) - Through (0,0): \( C=0 \), so **Particular solution:** \(\frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| = x \) - Through (0,2): \( y=2 \) is outside the domain of \(\tanh\); the solution is valid for \(|y|<1\). **Analytical solution:** \( y = \tanh(x + C) \), with \( C=0 \) for (0,0). --- ### **5c. \( y y' + x=0 \Rightarrow y' = -\frac{x}{y} \)** **Solution:** \( y^2 + x^2 = 2C \) - Through (0,1): \( 0^2 + 1^2=2C \Rightarrow C= \frac{1}{2} \) **Particular solution:** \(\boxed{ y^2 + x^2 = 1 } \) (a circle of radius 1 centered at (0,0))