Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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(](https://content.bartleby.com/qna-images/question/3d3733a5-5e9e-433b-b6d1-2cfec636672d/d04fefb4-fbcb-45bf-b4c8-1b6347139eca/kb3fjfx_processed.png)
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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