Check Cos(ne)= Cea (n(etsT) Peasoni Caslnle+am)) = Coa(ne +na) %3D
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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I don't understand why cos(ntheta+ n2pi)=cos(ntheta). Can you please explain it to me? Thank you
![## Overview of Differential Equation Consideration
### Equation and Condition
Consider the differential equation:
\[ S'' + \lambda S = 0 \]
with the periodic condition:
\[ S(\theta) = S(\theta + 2\pi) \]
### Case Analysis
1. **Case 1: \(\lambda = 0\)**
- General Solution:
\[ S(\theta) = C_1 + C_2 \theta \]
- Applying the periodic condition:
- Requires \( C_2 = 0 \), so
\[ S(\theta) = C_1 \]
2. **Case 2: \(\lambda = -\alpha^2 < 0\)**
- General Solution:
\[ S(\theta) = C_1 \cosh(\alpha \theta) + C_2 \sinh(\alpha \theta) \]
- This solution is not periodic, hence it is not considered further.
3. **Case 3: \(\lambda = \alpha^2 > 0\)**
- General Solution:
\[ S(\theta) = C_1 \cos(\alpha \theta) + C_2 \sin(\alpha \theta) \]
- The periodic condition \( S(\theta) = S(\theta + 2\pi) \) will hold if
\[ \alpha = n \]
- where \( n = 1, 2, 3, \ldots \) and \( n \in \mathbb{N} \).
### Verification
- Check using \( \cos(n\theta) = \cos(n(\theta + 2\pi)) \):
- \(\cos(n(\theta + 2\pi)) = \cos(n\theta + n \cdot 2\pi)\)
- \[ = \cos(n\theta) \]
This check confirms that the solution holds as expected for \(\alpha = n\).
This explanation provides a clear breakdown of each case for \(\lambda\) and ensures a thoughtful understanding of periodic solutions to the given differential equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F4fd8174a-d25f-46a5-9891-2c41187827cf%2Fylnr636_processed.png&w=3840&q=75)
Transcribed Image Text:## Overview of Differential Equation Consideration
### Equation and Condition
Consider the differential equation:
\[ S'' + \lambda S = 0 \]
with the periodic condition:
\[ S(\theta) = S(\theta + 2\pi) \]
### Case Analysis
1. **Case 1: \(\lambda = 0\)**
- General Solution:
\[ S(\theta) = C_1 + C_2 \theta \]
- Applying the periodic condition:
- Requires \( C_2 = 0 \), so
\[ S(\theta) = C_1 \]
2. **Case 2: \(\lambda = -\alpha^2 < 0\)**
- General Solution:
\[ S(\theta) = C_1 \cosh(\alpha \theta) + C_2 \sinh(\alpha \theta) \]
- This solution is not periodic, hence it is not considered further.
3. **Case 3: \(\lambda = \alpha^2 > 0\)**
- General Solution:
\[ S(\theta) = C_1 \cos(\alpha \theta) + C_2 \sin(\alpha \theta) \]
- The periodic condition \( S(\theta) = S(\theta + 2\pi) \) will hold if
\[ \alpha = n \]
- where \( n = 1, 2, 3, \ldots \) and \( n \in \mathbb{N} \).
### Verification
- Check using \( \cos(n\theta) = \cos(n(\theta + 2\pi)) \):
- \(\cos(n(\theta + 2\pi)) = \cos(n\theta + n \cdot 2\pi)\)
- \[ = \cos(n\theta) \]
This check confirms that the solution holds as expected for \(\alpha = n\).
This explanation provides a clear breakdown of each case for \(\lambda\) and ensures a thoughtful understanding of periodic solutions to the given differential equation.
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