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' = y - 5x, y(3) = 0. Y1 = Y₂ = Y3 = Y4=
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' = y - 5x, y(3) = 0. Y1 = Y₂ = Y3 = Y4=
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Using Euler's Method to Approximate Solutions of Differential Equations**
To apply Euler's method with a step size of 0.5 to compute the approximate \( y \)-values \( y_1, y_2, y_3 \), and \( y_4 \) for the initial-value problem:
\[
y' = y - 5x, \quad y(3) = 0
\]
we will follow these steps:
1. **Initialize the starting values:**
- \( x_0 = 3 \)
- \( y_0 = 0 \)
2. **Compute each subsequent \( y \)-value using the formula:**
\[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \]
where \( h = 0.5 \) is the step size and \( f(x, y) = y - 5x \).
### Iteration 1
\[
x_1 = x_0 + 0.5 = 3.5
\]
\[
y_1 = y_0 + 0.5 \cdot (y_0 - 5x_0) = 0 + 0.5 \cdot (0 - 15) = 0 - 7.5 = -7.5
\]
### Iteration 2
\[
x_2 = x_1 + 0.5 = 4.0
\]
\[
y_2 = y_1 + 0.5 \cdot (y_1 - 5x_1) = -7.5 + 0.5 \cdot (-7.5 - 17.5) = -7.5 + 0.5 \cdot (-25) = -7.5 - 12.5 = -20
\]
### Iteration 3
\[
x_3 = x_2 + 0.5 = 4.5
\]
\[
y_3 = y_2 + 0.5 \cdot (y_2 - 5x_2) = -20 + 0.5 \cdot (-20 - 20) = -20 + 0.5 \cdot (-40) = -20 - 20 = -40
\]
### Iteration 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad29c243-17bc-495f-9a75-f51d725c2e92%2F1d2cc913-7c65-492a-9357-39dec0633d11%2Fgg69g4d_processed.png&w=3840&q=75)
Transcribed Image Text:**Using Euler's Method to Approximate Solutions of Differential Equations**
To apply Euler's method with a step size of 0.5 to compute the approximate \( y \)-values \( y_1, y_2, y_3 \), and \( y_4 \) for the initial-value problem:
\[
y' = y - 5x, \quad y(3) = 0
\]
we will follow these steps:
1. **Initialize the starting values:**
- \( x_0 = 3 \)
- \( y_0 = 0 \)
2. **Compute each subsequent \( y \)-value using the formula:**
\[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \]
where \( h = 0.5 \) is the step size and \( f(x, y) = y - 5x \).
### Iteration 1
\[
x_1 = x_0 + 0.5 = 3.5
\]
\[
y_1 = y_0 + 0.5 \cdot (y_0 - 5x_0) = 0 + 0.5 \cdot (0 - 15) = 0 - 7.5 = -7.5
\]
### Iteration 2
\[
x_2 = x_1 + 0.5 = 4.0
\]
\[
y_2 = y_1 + 0.5 \cdot (y_1 - 5x_1) = -7.5 + 0.5 \cdot (-7.5 - 17.5) = -7.5 + 0.5 \cdot (-25) = -7.5 - 12.5 = -20
\]
### Iteration 3
\[
x_3 = x_2 + 0.5 = 4.5
\]
\[
y_3 = y_2 + 0.5 \cdot (y_2 - 5x_2) = -20 + 0.5 \cdot (-20 - 20) = -20 + 0.5 \cdot (-40) = -20 - 20 = -40
\]
### Iteration 4
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