c. y' + 2y = x³ with y(1) = 2 Yo (x) = e Yp (x) = y (x) = y (x) = -2x 1½ e ²x³ - 2/ ( 10²x² - (2x0²³²))+c 2-(2 3 2x 3 2x 2 X - ze 3 (4x²³ − 6x² + 6x + 8 ·C· e² (15e¯ -2x+2 +4x³ + 6x - 6x² - 3) -3) +C X
c. y' + 2y = x³ with y(1) = 2 Yo (x) = e Yp (x) = y (x) = y (x) = -2x 1½ e ²x³ - 2/ ( 10²x² - (2x0²³²))+c 2-(2 3 2x 3 2x 2 X - ze 3 (4x²³ − 6x² + 6x + 8 ·C· e² (15e¯ -2x+2 +4x³ + 6x - 6x² - 3) -3) +C X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:Solve the following IFOLDEs (for the specific solution as indicated) using the particular solution method. For each question you need to enter:
- The solution to the homogeneous equation; omit any multiplicative constant here.
- The particular solution with its constants determined.
- The general solution in the form \( C \cdot y_0(x) + y_p(x) \).
- The general solution after applying the given condition.
![### Transcription for Educational Website
**Problem Statement:**
Solve the differential equation: \( y' + 2y = x^3 \) with the initial condition \( y(1) = 2 \).
**General Solution for the Homogeneous Equation:**
\[ y_0(x) = e^{-2x} \]
**Particular Solution Attempt (Incorrect):**
\[ y_p(x) = \frac{1}{2} e^{2x} x^3 - \frac{3}{2} \left( \frac{1}{2} e^{2x} x^2 - \frac{1}{4} \left( 2xe^{2x} - e^{2x} \right) \right) + C \]
**Correct General Solution:**
\[ y(x) = \frac{1}{8} \left( 4x^3 - 6x^2 + 6x + 8 \cdot C \cdot e^{-2x} - 3 \right) \]
**Correct Particular Solution with Initial Condition:**
\[ y(x) = \frac{1}{8} \left( 15e^{-2x+2} + 4x^3 + 6x - 6x^2 - 3 \right) \]
**Explanation of Steps:**
1. **Solve the Homogeneous Equation:**
- The solution to \( y' + 2y = 0 \) is \( y_0(x) = e^{-2x} \).
2. **Propose a Particular Solution:**
- An incorrect form was attempted which did not satisfy the given equation or initial condition.
3. **Combine to Find the General Solution:**
- Combine homogeneous and particular solutions, adjusting constants to satisfy initial conditions.
4. **Apply Initial Condition:**
- Use \( y(1) = 2 \) to determine the constant in the solution.
Please note that the \( \times \) and \( \checkmark \) symbols indicate incorrect and correct steps, respectively, in the problem-solving process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8e0ad34-47bd-4edd-8e64-7fbb4be3cef6%2F1cccf786-805b-4d59-9852-aeabd7420d0d%2Fvdx4tw_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription for Educational Website
**Problem Statement:**
Solve the differential equation: \( y' + 2y = x^3 \) with the initial condition \( y(1) = 2 \).
**General Solution for the Homogeneous Equation:**
\[ y_0(x) = e^{-2x} \]
**Particular Solution Attempt (Incorrect):**
\[ y_p(x) = \frac{1}{2} e^{2x} x^3 - \frac{3}{2} \left( \frac{1}{2} e^{2x} x^2 - \frac{1}{4} \left( 2xe^{2x} - e^{2x} \right) \right) + C \]
**Correct General Solution:**
\[ y(x) = \frac{1}{8} \left( 4x^3 - 6x^2 + 6x + 8 \cdot C \cdot e^{-2x} - 3 \right) \]
**Correct Particular Solution with Initial Condition:**
\[ y(x) = \frac{1}{8} \left( 15e^{-2x+2} + 4x^3 + 6x - 6x^2 - 3 \right) \]
**Explanation of Steps:**
1. **Solve the Homogeneous Equation:**
- The solution to \( y' + 2y = 0 \) is \( y_0(x) = e^{-2x} \).
2. **Propose a Particular Solution:**
- An incorrect form was attempted which did not satisfy the given equation or initial condition.
3. **Combine to Find the General Solution:**
- Combine homogeneous and particular solutions, adjusting constants to satisfy initial conditions.
4. **Apply Initial Condition:**
- Use \( y(1) = 2 \) to determine the constant in the solution.
Please note that the \( \times \) and \( \checkmark \) symbols indicate incorrect and correct steps, respectively, in the problem-solving process.
Expert Solution
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