Compute y' and y". The symbols C₁ and C₂ represent constants. y = C₁ex sin x + C₂ex cos x y'(x) = y"(x) = Combine these derivatives with y as a linear second-order differential equation that is free of the symbols C₁ and C₂ and has the = 0 Need Help? Read It
Compute y' and y". The symbols C₁ and C₂ represent constants. y = C₁ex sin x + C₂ex cos x y'(x) = y"(x) = Combine these derivatives with y as a linear second-order differential equation that is free of the symbols C₁ and C₂ and has the = 0 Need Help? Read It
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
Compute \( y' \) and \( y'' \). The symbols \( C_1 \) and \( C_2 \) represent constants.
Given:
\[ y = C_1 e^x \sin x + C_2 e^x \cos x \]
Determine the following:
\[ y'(x) = \]
\[ y''(x) = \]
Combine these derivatives with \( y \) as a linear second-order differential equation that is free of the symbols \( C_1 \) and \( C_2 \) and has the form:
\[ \boxed{} = 0 \]
### Explanation of the Process
1. **First Derivative \( y' \):**
Apply the product rule to find \( y' \):
\[ y' = (C_1 e^x \sin x + C_2 e^x \cos x)' \]
\[ y' = C_1 \left( e^x \sin x \right)' + C_2 \left( e^x \cos x \right)' \]
For \( C_1 e^x \sin x \):
\[ \left( e^x \sin x \right)' = e^x \sin x + e^x \cos x \]
Thus,
\[ (C_1 e^x \sin x)' = C_1 \left( e^x \sin x + e^x \cos x \right) \]
\[ = C_1 e^x \sin x + C_1 e^x \cos x \]
For \( C_2 e^x \cos x \):
\[ \left( e^x \cos x \right)' = e^x \cos x - e^x \sin x \]
Thus,
\[ (C_2 e^x \cos x)' = C_2 \left( e^x \cos x - e^x \sin x \right) \]
\[ = C_2 e^x \cos x - C_2 e^x \sin x \]
Combining the results:
\[ y' = (C_1 e^x \sin x + C_1 e^x \cos x) + (C_2 e^x \cos x - C_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0436118d-47b7-4fa9-abd3-dac72bbeccd1%2F27599758-421a-47ae-9d8c-a2084adb6c2e%2F101glau_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Compute \( y' \) and \( y'' \). The symbols \( C_1 \) and \( C_2 \) represent constants.
Given:
\[ y = C_1 e^x \sin x + C_2 e^x \cos x \]
Determine the following:
\[ y'(x) = \]
\[ y''(x) = \]
Combine these derivatives with \( y \) as a linear second-order differential equation that is free of the symbols \( C_1 \) and \( C_2 \) and has the form:
\[ \boxed{} = 0 \]
### Explanation of the Process
1. **First Derivative \( y' \):**
Apply the product rule to find \( y' \):
\[ y' = (C_1 e^x \sin x + C_2 e^x \cos x)' \]
\[ y' = C_1 \left( e^x \sin x \right)' + C_2 \left( e^x \cos x \right)' \]
For \( C_1 e^x \sin x \):
\[ \left( e^x \sin x \right)' = e^x \sin x + e^x \cos x \]
Thus,
\[ (C_1 e^x \sin x)' = C_1 \left( e^x \sin x + e^x \cos x \right) \]
\[ = C_1 e^x \sin x + C_1 e^x \cos x \]
For \( C_2 e^x \cos x \):
\[ \left( e^x \cos x \right)' = e^x \cos x - e^x \sin x \]
Thus,
\[ (C_2 e^x \cos x)' = C_2 \left( e^x \cos x - e^x \sin x \right) \]
\[ = C_2 e^x \cos x - C_2 e^x \sin x \]
Combining the results:
\[ y' = (C_1 e^x \sin x + C_1 e^x \cos x) + (C_2 e^x \cos x - C_
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