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
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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_
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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