Let's solve each problem **step by step**, focusing on **how to construct the direction field** and **sketch particular solutions**. I'll provide the method and the analytic solution where possible. --- ## **5a) \( y' = 1 + y^2 \), Point: (1,1)** ### **Step 1: Understand the Equation** - \( y' = 1 + y^2 \) - This is a first-order, nonlinear ODE. - The slope at each point depends only on \( y \). ### **Step 2: Create the Direction Field** - For various points \((x, y)\) within the plot limits, calculate the slope \( y' \). - **Resolution:** 0.5 (so plot at \( x = 0, 0.5, 1, ... \), \( y = -3, -2.5, ..., 3 \)) - **Slope computation:** At each point, slope is \( 1 + y^2 \) (does not depend on \( x \)). #### **Examples:** - At \( y = 0 \): \( y' = 1 \) - At \( y = 1 \): \( y' = 2 \) - At \( y = -1 \): \( y' = 2 \) - At \( y = 2 \): \( y' = 5 \) - At \( y = -2 \): \( y' = 5 \) - At \( y = 3 \): \( y' = 10 \) So, as \( |y| \) increases, the slope gets steeper. ### **Step 3: Sketch the Particular Solution** - **Through (1,1):** This is the initial condition. - 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 (1,1): \( \arctan(1) = 1 + C \implies \frac{\pi}{4} = 1 + C \implies C = \frac{\pi}{4} - 1 \) So, \[ \arctan(y) = x + \frac{\pi}{4} - 1 \] Or, \[ y(x) = \tan \left( x + \frac{\pi}{4} - 1 \right) \] ### **Step 4: Final Answer** - Draw the direction field with slopes as above. - Draw the particular solution curve through (1,1) using the formula above. --- ## **5b) \( y' = 1 - y^2 \), Points: (0,0), (0,2)** ### **Step 1: Understand the Equation** - Slope \( y' = 1 - y^2 \) ### **Step 2: Create the Direction Field** - **Resolution:** 0.5 - **Slope computation:** - At \( y = 0 \): \( y' = 1 \) - At \( y = 1 \): \( y' = 0 \) - At \( y = -1 \): \( y' = 0 \) - At \( y = 2 \): \( y' = -3 \) - At \( y = -2 \): \( y' = -3 \) ### **Step 3: Sketch the Particular Solutions** - **Through (0,0):** \[ \frac{dy}{dx} = 1 - y^2 \implies \int \frac{1}{1 - y^2} dy = \int dx \] \[ \int \left( \frac{1}{2} \left( \frac{1}{1-y} + \frac{1}{1+y} \right) \right) dy = x + C \] \[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x + C \] Using (0,0): \( C = 0 \) \[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x \] \[ \frac{1+y}{1-y} = e^{2x} \] \[ y = \frac{e^{2x} - 1}{e^{2x} + 1} \] - **Through (0,2):** \[ \frac{1}{2} \ln \left| \frac{1+2}{1-2} \right| = C \implies \frac{1}{2} \ln \left| \frac{3}{-1} \right| = C \implies \frac{1}{2} \ln 3 - \frac{1}{2} \ln 1 = \frac{1}{2} \ln 3 \] So, \[ x + C = x + \frac{1}{2} \ln 3 \] Or, \[ \frac{1}{2} \ln \left| \frac{1+y}{1-y} \right| = x + \frac{1}{2} \ln 3 \] \[ \ln \left| \frac{1+y}{1-y} \right| = 2x + \ln 3 \] \[ \frac{1+y}{1-y} = 3 e^{2x} \] \[ y = \frac{3e^{2x} - 1}{3e^{2x} + 1} \] ### **Step 4: Final Answer** - Draw the direction field as above. - Draw two solution curves: one through (0,0), one through (0,2) with the derived formulas. --- ## **5c) \( y y' + x = 0 \), Point: (0,1)** ### **Step 1: Understand the Equation** - \( y y' = -x \implies y' = -x/y \) ### **Step 2: Create the Direction Field** - **Resolution:** 0.5 - **Slope computation:** At each (x, y), slope is \( -x/y \). - At \( x = 0 \): \( y' = 0 \) (horizontal slopes along y-axis) - At \( y = 0 \): undefined (vertical asymptote) ### **Step 3: Sketch the Particular Solution** - **Separate variables:** \[ y \frac{dy}{dx} = -x \implies y dy = -x dx \] \[ \int y dy = -\int x dx \implies \frac{1}{2} y^2 = -\frac{1}{2} x^2 + C \] Using (0,1): \( \frac{1}{2} (1)^2 = C \implies C = \frac{1}{2} \) So, \[ \frac{1}{2} y^2 = -\frac{1}{2} x^2 + \frac{1}{2} \implies y^2 + x^2 = 1 \implies y = \sqrt{1 - x^2} \] (Take the positive root for \( y > 0 \)) ### **Step 4: Final Answer** - Draw the direction field using the slopes. - Draw the solution curve \( y = \sqrt{1 - x^2} \) through (0,1). --- ## **Summary Table** | Problem | Differential Equation | Points | Particular Solution (Formula) | |---------|----------------------|--------|-------------------------------| | 5a | \( y' = 1 + y^2 \) | (1,1) | \( y(x) = \tan(x + \frac{\pi}{4} - 1) \) | | 5b | \( y' = 1 - y^2 \) | (0,0), (0,2) | \( y(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \), \( y(x) = \frac{3 e^{2x} - 1}{3 e^{2x} + 1} \) | | 5c | \( y y' + x = 0 \) | (0,1) | \( y(x) = \sqrt{1 - x^2} \) | --- If you need **sketches or direction field diagrams**, let me know!

Certainly! Below is a detailed, step-by-step breakdown for each problem, including calculations, explanations, and final solutions. This will guide you on how to graph the direction fields by hand and sketch particular solutions passing through the specified points. --- ## **Problem 5a) \( y' = 1 + y^2 \), Point: (1,1)** ### **Step 1: Understand the differential equation** - \( y' = 1 + y^2 \) - Slope depends only on \( y \), independent of \( x \). - As \( y \) increases, the slope \( y' \) increases quadratically. ### **Step 2: Set up a table of slopes for key \( y \) values** Choose \( y \)-values within the plot limits (say, from -3 to 3). Compute \( y' = 1 + y^2 \): | \( y \) | \( y' = 1 + y^2 \) | Slope value | |---------|---------------------|--------------| | -3 | \( 1 + 9 = 10 \) | Very steep downward (since \( y \) is negative, but slope is positive) | | -2 | \( 1 + 4 = 5 \) | Steep downward | | -1 | \( 1 + 1 = 2 \) | Moderate slope | | 0 | \( 1 + 0 = 1 \) | Gentle slope | | 1 | \( 1 + 1 = 2 \) | Same as \( y = -1 \) | | 2 | \( 1 + 4 = 5 \) | Steep slope | | 3 | \( 1 + 9 = 10 \) | Very steep slope | **Plot the direction vectors (small line segments) at each grid point** with the corresponding slopes. ### **Step 3: Find the particular solution passing through (1,1)** Because the equation is separable: \[ \frac{dy}{dx} = 1 + y^2 \] Separate variables: \[ \int \frac{dy}{1 + y^2} = \int dx \] Recall: \[ \int \frac{dy}{1 + y^2} = \arctan(y) + C \] So, \[ \arctan(y) = x + C \] Use the initial point \( (x, y) = (1, 1) \): \[ \arctan(1) = 1 + C \implies \frac{\pi}{4} = 1 + C \implies C = \frac{\pi}{4} - 1 \] Final particular solution: \[ \boxed{ \arctan(y) = x + \frac{\pi}{4} - 1 } \] or equivalently, \[ \boxed{ y(x) = \tan \left( x + \frac{\pi}{4} - 1 \right) } \] ### **Step 4: Sketch the solution** - Plot the direction vectors to visualize the slope field. - Sketch the curve \( y(x) \) passing through (1,1): - At \( x=1 \), \[ y(1) = \tan \left( 1 + \frac{\pi}{4} - 1 \right) = \tan \left( \frac{\pi}{4} \right) = 1 \] - For \( x \) near 1, the curve follows the tangent of the shifted \( x \). - As \( x \) approaches the vertical asymptote \( \arctan(y) = \pm \frac{\pi}{2} \), the solutions tend to infinity. --- ## **Problem 5b) \( y' = 1 - y^2 \), Points: (0,0) and (0,2)** ### **Step 1: Understand the differential equation** - \( y' = 1 - y^2 \) - This is a Bernoulli-type nonlinear ODE. - Slope depends only on \( y \). ### **Step 2: Compute slopes at key \( y \)-values** | \( y \) | \( y' = 1 - y^2 \) | Slope | |---------|---------------------|--------| | -2 | \( 1 - 4 = -3 \) | steep downward | | -1 | \( 1 - 1 = 0 \) | horizontal | | 0 | \( 1 - 0 = 1 \) | gentle upward | | 1 | \( 1 - 1 = 0 \) | horizontal | | 2 | \( 1 - 4 = -3 \) | steep downward | ### **Step 3: Find particular solutions passing through the points** --- **(i) Passing through (0,0):** Separate variables: \[ \frac{dy}{dx} = 1 - y^2 \] \[ \int \frac{dy}{1 - y^2} = \int dx \] Recall the integral: \[ \int \frac{dy}{1 - y^2} = \frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| + C \] Set: \[ \frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| = x + C \] Use initial point \( (0,0) \): \[ \frac{1}{2} \ln \left| \frac{1 + 0}{1 - 0} \right| = 0 + C \Rightarrow \frac{1}{2} \ln 1 = C \Rightarrow C = 0 \] So, \[ \frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| = x \] Solve for \( y \): \[ \left| \frac{1 + y}{1 - y} \right| = e^{2x} \] Assuming \( y \) in the range \(-1 < y < 1 \): \[ \frac{1 + y}{1 - y} = e^{2x} \] Finally, \[ y(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \] --- **(ii) Passing through (0,2):** Initial condition: \[ \frac{1}{2} \ln \left| \frac{1 + 2}{1 - 2} \right| = C \] Calculate: \[ \frac{1}{2} \ln \left| \frac{3}{-1} \right| = \frac{1}{2} \ln 3 = C \] The solution: \[ \frac{1}{2} \ln \left| \frac{1 + y}{1 - y} \right| = x + C = x + \frac{1}{2} \ln 3 \] Exponentiating: \[ \left| \frac{1 + y}{1 - y} \right| = e^{2x + \ln 3} = 3 e^{2x} \] Assuming \( y \) in \(-1 < y < 1 \): \[ \frac{1 + y}{1 - y} = 3 e^{2x} \] Solve for \( y \): \[ 1 + y = 3 e^{2x} (1 - y) \] \[ 1 + y = 3 e^{2x} - 3 e^{2x} y \] \[ y + 3 e^{2x} y = 3 e^{2x} - 1 \] \[ y (1 + 3 e^{2x}) = 3 e^{2x} - 1 \] \[ \boxed{ y(x) = \frac{3 e^{2x} - 1}{1 + 3 e^{2x}} } \] --- ### **Step 4: Sketch the direction field and particular solutions** - Plot slopes at grid points using the values above. - Draw the two solution curves: - \( y = \frac{e^{2x} - 1}{e^{2x} + 1} \) passing through (0,0). - \( y = \frac{3 e^{2x} - 1}{1 + 3 e^{2x}} \) passing through (0,2). --- ## **Problem 5c) \( y y' + x = 0 \), Point: (0,1)** ### **Step 1: Understand the equation** \[ y y' + x = 0 \] \[ \Rightarrow y y' = -x \] \[ \Rightarrow y' = -\frac{x}{y} \] - Slope at \( (x, y) \): \( -x/y \). ### **Step 2: Create the slope table** At various points: | \( x \) | \( y \) | \( y' = -x/y \) | |---------|----------|----------------| | 0 | 1 | 0 | | 0.5 | 1 | \( -0.5/1 = -0.5 \) | | 1 | 1 | \( -1/1 = -1 \) | | 0 | 2 | \( 0/2 = 0 \) | | 1 | 2 | \( -1/2 = -0.5 \) | | 2 | 1 | \( -2/1 = -2 \) | Note: Along \( x=0 \), slopes are zero, indicating horizontal tangents. ### **Step 3: Find the particular solution passing through (0,1)** Separate variables: \[ y y' = -x \] \[ \Rightarrow y dy = -x dx \] Integrate both sides: \[ \int y dy = - \int x dx \] \[ \frac{1}{2} y^2 = - \frac{1}{2} x^2 + C \] Apply initial condition \( (0,1) \): \[ \frac{1}{2} (1)^2 = - \frac{1}{2} (0)^2 + C \Rightarrow \frac{1}{2} = C \] Thus, \[ \frac{1}{2} y^2 = - \frac{1}{2} x^2 + \frac{1}{2} \] Multiply through by 2: \[ y^2 = -x^2 + 1 \] \[ \boxed{ y = \pm \sqrt{1 - x^2} } \] Since the point is (0,1), the particular solution is: \[ \boxed{ y = \sqrt{1 - x^2} } \] This is a semicircle of radius 1 centered at the origin. ### **Step 4: Sketch the solution** - The solution curve is the upper semicircle \( y = \sqrt{1 - x^2} \). - The slope field along \( x=0 \) is horizontal (slope 0), matching the curve's tangent at \( (0,1) \). - The direction vectors will point along the semicircle, with slopes given by \( -x/y \). --- # **Summary of Final Answers** | Problem | Differential Equation | Particular Solution | Remarks | |-----------|------------------------|------------------------|---------| | **5a** | \( y' = 1 + y^2 \) | \( y(x) = \tan \left( x + \frac{\pi}{4} - 1 \right) \) | Vertical asymptote where tangent blows up. | | **5b** | \( y' = 1 - y^2 \) | \( y(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \) | Solution through (0,0); similar for (0,2). | | **5c** | \( y y' + x = 0 \) | \( y = \sqrt{1 - x^2} \) | Semicircular solution passing through (0,1). | --- ### **How to graph by hand:** - Plot a grid of points. - For each, compute the slope using the equations above. - Draw small line segments at each point with these slopes to visualize the direction field. - Use the particular solutions formulas to sketch the specific curves passing through the given points. If you'd like, I can also guide you on how to sketch the direction fields or provide more detailed plotting tips!