Use Euler's method with step size 0.3 to compute the approximate y-values y(1.3) and y(1.6), of the solution of the initial-value problem y' = 1 – 3x + 4y, y(1) = - 3.
Use Euler's method with step size 0.3 to compute the approximate y-values y(1.3) and y(1.6), of the solution of the initial-value problem y' = 1 – 3x + 4y, y(1) = - 3.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**Euler's Method for Approximating Solutions to Differential Equations**
To illustrate the use of Euler's method, consider the initial-value problem described by the differential equation:
\[ y' = 1 - 3x + 4y, \quad y(1) = -3. \]
We are tasked with computing the approximate \( y \)-values \( y(1.3) \) and \( y(1.6) \) using a step size of 0.3.
Euler's method uses the formula:
\[ y_{n+1} = y_n + h f(x_n, y_n) \]
where \( h \) is the step size, and \( f(x,y) \) represents the function on the right-hand side of the differential equation.
**Step-by-Step Solution:**
1. **Initial Values:**
- \( x_0 = 1 \)
- \( y_0 = -3 \)
2. **Step Size:**
- \( h = 0.3 \)
3. **First Iteration (to find \( y(1.3) \)):**
- Calculate the slope at the initial point: \( f(x_0, y_0) = 1 - 3(1) + 4(-3) = 1 - 3 - 12 = -14 \).
- Apply Euler's formula: \( y_1 = y_0 + h f(x_0, y_0) = -3 + 0.3(-14) = -3 - 4.2 = -7.2 \).
- So, \( y(1.3) \approx -7.2 \).
4. **Second Iteration (to find \( y(1.6) \)):**
- Update \( x \) to 1.3 and \( y \) to -7.2.
- Calculate the new slope: \( f(x_1, y_1) = 1 - 3(1.3) + 4(-7.2) = 1 - 3.9 - 28.8 = 1 - 32.7 = -31.7 \).
- Apply Euler's formula again: \( y_2 = y_1 + h f(x_1, y_1) = -7.2 + 0.3(-31](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62b092f3-63db-4d93-982f-67d0473d8e68%2F330911b6-9e35-4058-a77b-ec58470b527f%2Fnjtjp83_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Euler's Method for Approximating Solutions to Differential Equations**
To illustrate the use of Euler's method, consider the initial-value problem described by the differential equation:
\[ y' = 1 - 3x + 4y, \quad y(1) = -3. \]
We are tasked with computing the approximate \( y \)-values \( y(1.3) \) and \( y(1.6) \) using a step size of 0.3.
Euler's method uses the formula:
\[ y_{n+1} = y_n + h f(x_n, y_n) \]
where \( h \) is the step size, and \( f(x,y) \) represents the function on the right-hand side of the differential equation.
**Step-by-Step Solution:**
1. **Initial Values:**
- \( x_0 = 1 \)
- \( y_0 = -3 \)
2. **Step Size:**
- \( h = 0.3 \)
3. **First Iteration (to find \( y(1.3) \)):**
- Calculate the slope at the initial point: \( f(x_0, y_0) = 1 - 3(1) + 4(-3) = 1 - 3 - 12 = -14 \).
- Apply Euler's formula: \( y_1 = y_0 + h f(x_0, y_0) = -3 + 0.3(-14) = -3 - 4.2 = -7.2 \).
- So, \( y(1.3) \approx -7.2 \).
4. **Second Iteration (to find \( y(1.6) \)):**
- Update \( x \) to 1.3 and \( y \) to -7.2.
- Calculate the new slope: \( f(x_1, y_1) = 1 - 3(1.3) + 4(-7.2) = 1 - 3.9 - 28.8 = 1 - 32.7 = -31.7 \).
- Apply Euler's formula again: \( y_2 = y_1 + h f(x_1, y_1) = -7.2 + 0.3(-31
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