Are the functions y₁(x) = e, y2(x) = ze a fundamental set of solutions for the equation y" - 2y + y = 0? Justify your answer. If yes, write the general solution of y" - 2y + y = 0.
Are the functions y₁(x) = e, y2(x) = ze a fundamental set of solutions for the equation y" - 2y + y = 0? Justify your answer. If yes, write the general solution of y" - 2y + y = 0.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
"Are the functions \( y_1(x) = e^x \), \( y_2(x) = xe^x \) a fundamental set of solutions for the equation \( y'' - 2y' + y = 0 \)? Justify your answer. If yes, write the general solution of \( y'' - 2y' + y = 0 \)."
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**Detailed Solution Explanation:**
To determine if \( y_1(x) = e^x \) and \( y_2(x) = xe^x \) form a fundamental set of solutions for the differential equation \( y'' - 2y' + y = 0 \), we need to show two things:
1. Both \( y_1(x) \) and \( y_2(x) \) are solutions to the differential equation.
2. \( y_1(x) \) and \( y_2(x) \) are linearly independent.
**Step 1: Verify Solutions**
1. Taking derivatives of \( y_1(x) = e^x \):
- \( y_1' = e^x \)
- \( y_1'' = e^x \)
Substitute into \( y'' - 2y' + y = 0 \):
\[
e^x - 2e^x + e^x = 0 \implies 0 = 0
\]
Therefore, \( y_1(x) = e^x \) is a solution.
2. Taking derivatives of \( y_2(x) = xe^x \):
- \( y_2' = e^x + xe^x = (1 + x)e^x \)
- \( y_2'' = (1 + x)e^x + e^x = (2 + x)e^x \)
Substitute into \( y'' - 2y' + y = 0 \):
\[
(2 + x)e^x - 2(1 + x)e^x + xe^x = 0 \implies (2 + x)e^x - 2e^x - 2xe^x + xe^x = 0 \implies 0 = 0
\]
Therefore, \( y_2(x) = xe^x \) is also a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc47a605b-1167-4b41-bc40-883b330bf596%2F56272d5f-7891-40f7-8f62-14c3ab280383%2Fgni8bgb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
"Are the functions \( y_1(x) = e^x \), \( y_2(x) = xe^x \) a fundamental set of solutions for the equation \( y'' - 2y' + y = 0 \)? Justify your answer. If yes, write the general solution of \( y'' - 2y' + y = 0 \)."
---
**Detailed Solution Explanation:**
To determine if \( y_1(x) = e^x \) and \( y_2(x) = xe^x \) form a fundamental set of solutions for the differential equation \( y'' - 2y' + y = 0 \), we need to show two things:
1. Both \( y_1(x) \) and \( y_2(x) \) are solutions to the differential equation.
2. \( y_1(x) \) and \( y_2(x) \) are linearly independent.
**Step 1: Verify Solutions**
1. Taking derivatives of \( y_1(x) = e^x \):
- \( y_1' = e^x \)
- \( y_1'' = e^x \)
Substitute into \( y'' - 2y' + y = 0 \):
\[
e^x - 2e^x + e^x = 0 \implies 0 = 0
\]
Therefore, \( y_1(x) = e^x \) is a solution.
2. Taking derivatives of \( y_2(x) = xe^x \):
- \( y_2' = e^x + xe^x = (1 + x)e^x \)
- \( y_2'' = (1 + x)e^x + e^x = (2 + x)e^x \)
Substitute into \( y'' - 2y' + y = 0 \):
\[
(2 + x)e^x - 2(1 + x)e^x + xe^x = 0 \implies (2 + x)e^x - 2e^x - 2xe^x + xe^x = 0 \implies 0 = 0
\]
Therefore, \( y_2(x) = xe^x \) is also a
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