Use Euler's method to compute the approximate y-values, y(1.2) and y(1.4), of the solution of the initial- value problem y' = 1 − 4x + 2y, y(1) = - 2. y(1.2) = y(1.4) =

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### Euler's Method for Approximation

Euler's method is a numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. It's a first-order method, meaning it approximates the solution by considering the slope of the function at discrete points. 

Below is a problem that involves using Euler's method to compute approximate values of \( y \) for a given initial-value problem.

#### Problem Statement

Use Euler’s method to compute the approximate \( y \)-values, \( y(1.2) \) and \( y(1.4) \), of the solution of the initial-value problem:

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

#### Step-by-Step Process:

1. **Identify the differential equation and initial condition:**
   \[
   y' = 1 - 4x + 2y, \quad y(1) = -2.
   \]

2. **Choose a step size \( h \).** (Let's assume \( h = 0.2 \).)

3. **Compute \( y(1.2) \):**
   \[
   y(1.2) = y(1) + h \cdot f(x, y)
   \]
   where \( f(x, y) = 1 - 4x + 2y \).

   Given \( x = 1 \) and \( y = -2 \):
   \[
   f(1, -2) = 1 - 4(1) + 2(-2) = 1 - 4 - 4 = -7.
   \]
   Thus,
   \[
   y(1.2) = -2 + 0.2 \cdot (-7) = -2 - 1.4 = -3.4.
   \]

3. **Compute \( y(1.4) \):**
   Now, using the new point \( (1.2, -3.4) \):
   \[
   f(1.2, -3.4) = 1 - 4(1.2) + 2(-3.4) = 1 - 4.8 - 6.8 = -10.6.
   \]
   Thus,
   \[
   y(1.4) = y(
Transcribed Image Text:### Euler's Method for Approximation Euler's method is a numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. It's a first-order method, meaning it approximates the solution by considering the slope of the function at discrete points. Below is a problem that involves using Euler's method to compute approximate values of \( y \) for a given initial-value problem. #### Problem Statement Use Euler’s method to compute the approximate \( y \)-values, \( y(1.2) \) and \( y(1.4) \), of the solution of the initial-value problem: \[ y' = 1 - 4x + 2y, \quad y(1) = -2. \] #### Step-by-Step Process: 1. **Identify the differential equation and initial condition:** \[ y' = 1 - 4x + 2y, \quad y(1) = -2. \] 2. **Choose a step size \( h \).** (Let's assume \( h = 0.2 \).) 3. **Compute \( y(1.2) \):** \[ y(1.2) = y(1) + h \cdot f(x, y) \] where \( f(x, y) = 1 - 4x + 2y \). Given \( x = 1 \) and \( y = -2 \): \[ f(1, -2) = 1 - 4(1) + 2(-2) = 1 - 4 - 4 = -7. \] Thus, \[ y(1.2) = -2 + 0.2 \cdot (-7) = -2 - 1.4 = -3.4. \] 3. **Compute \( y(1.4) \):** Now, using the new point \( (1.2, -3.4) \): \[ f(1.2, -3.4) = 1 - 4(1.2) + 2(-3.4) = 1 - 4.8 - 6.8 = -10.6. \] Thus, \[ y(1.4) = y(
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