Find L {&* sin J61 -r’e * } -3t |
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![## Laplace Transform of a Given Function
### Problem Statement
Find the Laplace transform \( \mathcal{L} \{ e^{-3t} \sin (\sqrt{6} t) - t^3 e^{-3t} \} \).
### Explanation
The expression inside the Laplace transform operator \( \mathcal{L} \) consists of two terms:
1. \( e^{-3t} \sin (\sqrt{6} t) \)
2. \( - t^3 e^{-3t} \)
We aim to find the Laplace transform of these terms individually and then combine the results.
**Note**: The Laplace transform is an integral transform widely used to convert a function of time \( t \) to a function of complex frequency \( s \). The transform is especially useful for solving differential equations and analyzing linear time-invariant systems.
### Steps to Compute the Laplace Transform
1. **Laplace Transform of \( e^{-3t} \sin(\sqrt{6} t) \)**:
- The general form for the Laplace transform of \( e^{at} \sin(bt) \) is \( \mathcal{L}\{ e^{at} \sin(bt) \} = \frac{b}{(s-a)^2 + b^2} \).
- Here, \( a = -3 \) and \( b = \sqrt{6} \).
- Substituting these values into the formula gives us:
\[
\mathcal{L} \{ e^{-3t} \sin(\sqrt{6} t) \} = \frac{\sqrt{6}}{(s + 3)^2 + 6}
\]
2. **Laplace Transform of \( -t^3 e^{-3t} \)**:
- The general form for the Laplace transform of \( t^n e^{at} \) is \( \mathcal{L} \{ t^n e^{at} \} = \frac{n!}{(s-a)^{n+1}} \).
- Here, \( a = -3 \) and \( n = 3 \).
- Substituting these values into the formula gives us:
\[
\mathcal{L} \{ t^3 e^{-3t} \} = \frac{3!}{(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24a60763-065d-4deb-b582-ca07220062f2%2Fbc590a4f-82c0-4079-90e8-a2fd765bdedb%2Fkncngtm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Laplace Transform of a Given Function
### Problem Statement
Find the Laplace transform \( \mathcal{L} \{ e^{-3t} \sin (\sqrt{6} t) - t^3 e^{-3t} \} \).
### Explanation
The expression inside the Laplace transform operator \( \mathcal{L} \) consists of two terms:
1. \( e^{-3t} \sin (\sqrt{6} t) \)
2. \( - t^3 e^{-3t} \)
We aim to find the Laplace transform of these terms individually and then combine the results.
**Note**: The Laplace transform is an integral transform widely used to convert a function of time \( t \) to a function of complex frequency \( s \). The transform is especially useful for solving differential equations and analyzing linear time-invariant systems.
### Steps to Compute the Laplace Transform
1. **Laplace Transform of \( e^{-3t} \sin(\sqrt{6} t) \)**:
- The general form for the Laplace transform of \( e^{at} \sin(bt) \) is \( \mathcal{L}\{ e^{at} \sin(bt) \} = \frac{b}{(s-a)^2 + b^2} \).
- Here, \( a = -3 \) and \( b = \sqrt{6} \).
- Substituting these values into the formula gives us:
\[
\mathcal{L} \{ e^{-3t} \sin(\sqrt{6} t) \} = \frac{\sqrt{6}}{(s + 3)^2 + 6}
\]
2. **Laplace Transform of \( -t^3 e^{-3t} \)**:
- The general form for the Laplace transform of \( t^n e^{at} \) is \( \mathcal{L} \{ t^n e^{at} \} = \frac{n!}{(s-a)^{n+1}} \).
- Here, \( a = -3 \) and \( n = 3 \).
- Substituting these values into the formula gives us:
\[
\mathcal{L} \{ t^3 e^{-3t} \} = \frac{3!}{(
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