Use Euler's method with step size 0.5 to compute the approximate y-values y1, Y2, Y3, and y4 of the solution of the initial-value problem y' = - 2 + 2x + 4y, y(1) = 2. Y1 = Y2 = 93 = Y4 = I| || || ||

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### Using Euler's Method to Approximate Solutions to Differential Equations

Euler's method is a simple numerical approach to approximating solutions to ordinary differential equations (ODEs). Given the step size \( h \) and an initial value problem, the method estimates the values of the unknown function at discrete points.

#### Problem Statement
Use Euler's method with step size \( 0.5 \) to compute the approximate y-values \( y_1 \), \( y_2 \), \( y_3 \), and \( y_4 \) for the solution of the given initial-value problem:

\[ y' = -2 + 2x + 4y, \quad y(1) = 2. \]

#### Procedure

1. **Initial Value**:
   - \( x_0 = 1 \)
   - \( y_0 = 2 \)

2. **Step Size**:
   - \( h = 0.5 \)

3. **Equations**:
   \( y' = f(x, y) = -2 + 2x + 4y \)

4. **Iteration Formula**:
   \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \]

Use the iteration formula to compute \( y_1, y_2, y_3, \) and \( y_4 \).

#### Solution

1. **First Iteration**:
   - \( x_0 = 1 \), \( y_0 = 2 \)
   - Compute \( f(x_0, y_0) \):
     \[ f(1, 2) = -2 + 2(1) + 4(2) = -2 + 2 + 8 = 8 \]
   - Calculate \( y_1 \):
     \[ y_1 = y_0 + h \cdot f(x_0, y_0) = 2 + 0.5 \cdot 8 = 2 + 4 = 6 \]

2. **Second Iteration**:
   - \( x_1 = 1.5 \), \( y_1 = 6 \)
   - Compute \( f(x_1, y_1) \):
     \[ f(1.5, 6) = -2 + 2(1.5) +
Transcribed Image Text:### Using Euler's Method to Approximate Solutions to Differential Equations Euler's method is a simple numerical approach to approximating solutions to ordinary differential equations (ODEs). Given the step size \( h \) and an initial value problem, the method estimates the values of the unknown function at discrete points. #### Problem Statement Use Euler's method with step size \( 0.5 \) to compute the approximate y-values \( y_1 \), \( y_2 \), \( y_3 \), and \( y_4 \) for the solution of the given initial-value problem: \[ y' = -2 + 2x + 4y, \quad y(1) = 2. \] #### Procedure 1. **Initial Value**: - \( x_0 = 1 \) - \( y_0 = 2 \) 2. **Step Size**: - \( h = 0.5 \) 3. **Equations**: \( y' = f(x, y) = -2 + 2x + 4y \) 4. **Iteration Formula**: \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \] Use the iteration formula to compute \( y_1, y_2, y_3, \) and \( y_4 \). #### Solution 1. **First Iteration**: - \( x_0 = 1 \), \( y_0 = 2 \) - Compute \( f(x_0, y_0) \): \[ f(1, 2) = -2 + 2(1) + 4(2) = -2 + 2 + 8 = 8 \] - Calculate \( y_1 \): \[ y_1 = y_0 + h \cdot f(x_0, y_0) = 2 + 0.5 \cdot 8 = 2 + 4 = 6 \] 2. **Second Iteration**: - \( x_1 = 1.5 \), \( y_1 = 6 \) - Compute \( f(x_1, y_1) \): \[ f(1.5, 6) = -2 + 2(1.5) +
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