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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.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter5: Graphs And The Derivative
Section5.CR: Chapter 5 Review
Problem 16CR
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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 \( y'' - 2y' + y = 0 \)." --- **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
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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