Determine the inverse Laplace transform of the function below. 16 s² +25 Click here to view the table of Laplace transforms
Determine the inverse Laplace transform of the function below. 16 s² +25 Click here to view the table of Laplace transforms
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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![## Inverse Laplace Transform Problem
### Problem Statement:
Determine the inverse Laplace transform of the function below.
\[
\frac{16}{s^2 + 25}
\]
#### Interactive Resources:
- [Click here to view the table of Laplace transforms.](#)
- [Click here to view the table of properties of Laplace transforms.](#)
### Solution:
Utilize the inverse Laplace transform operator:
\[
\mathcal{L}^{-1}\left\{\frac{16}{s^2 + 25}\right\} = \Box
\]
---
## Properties of Laplace Transforms
### List of Properties:
1. \(\mathcal{L}\{f + g\} = \mathcal{L}\{f\} + \mathcal{L}\{g\}\)
2. \(\mathcal{L}\{cf\} = c\mathcal{L}\{f\}\) for any constant \(c\)
3. \(\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s - a)\)
4. \(\mathcal{L}\{f'(t)\}(s) = s\mathcal{L}\{f(t)\}(s) - f(0)\)
5. \(\mathcal{L}\{f''(t)\}(s) = s^2\mathcal{L}\{f(t)\}(s) - sf(0) - f'(0)\)
6. \(\mathcal{L}\{f^{(n)}(t)\}(s) = s^n\mathcal{L}\{f(t)\}(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - f^{(n-1)}(0)\)
7. \(\mathcal{L}\{f^{(n)}(t)\} = (-1)^n \frac{d^n}{ds^n} \{\mathcal{L}\{f(t)\}\}\)
8. \(\mathcal{L}^{-1} \{F_1 + F_2\} = \mathcal{L}^{-1} \{F_1\} + \mathcal{L](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F047a7e6a-f025-4b5b-ab83-4ffe14f69253%2F45896c8b-e58e-4bf5-8182-af1561f51df6%2Fgdb0onv_processed.png&w=3840&q=75)
Transcribed Image Text:## Inverse Laplace Transform Problem
### Problem Statement:
Determine the inverse Laplace transform of the function below.
\[
\frac{16}{s^2 + 25}
\]
#### Interactive Resources:
- [Click here to view the table of Laplace transforms.](#)
- [Click here to view the table of properties of Laplace transforms.](#)
### Solution:
Utilize the inverse Laplace transform operator:
\[
\mathcal{L}^{-1}\left\{\frac{16}{s^2 + 25}\right\} = \Box
\]
---
## Properties of Laplace Transforms
### List of Properties:
1. \(\mathcal{L}\{f + g\} = \mathcal{L}\{f\} + \mathcal{L}\{g\}\)
2. \(\mathcal{L}\{cf\} = c\mathcal{L}\{f\}\) for any constant \(c\)
3. \(\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s - a)\)
4. \(\mathcal{L}\{f'(t)\}(s) = s\mathcal{L}\{f(t)\}(s) - f(0)\)
5. \(\mathcal{L}\{f''(t)\}(s) = s^2\mathcal{L}\{f(t)\}(s) - sf(0) - f'(0)\)
6. \(\mathcal{L}\{f^{(n)}(t)\}(s) = s^n\mathcal{L}\{f(t)\}(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - f^{(n-1)}(0)\)
7. \(\mathcal{L}\{f^{(n)}(t)\} = (-1)^n \frac{d^n}{ds^n} \{\mathcal{L}\{f(t)\}\}\)
8. \(\mathcal{L}^{-1} \{F_1 + F_2\} = \mathcal{L}^{-1} \{F_1\} + \mathcal{L
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