Determine L'{F}. s+1 SF(s) – 3F(s) = - 2 s + 6s + 9 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. L¯{F} =U 1 {}, _B
Determine L'{F}. s+1 SF(s) – 3F(s) = - 2 s + 6s + 9 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. L¯{F} =U 1 {}, _B
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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Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question
![**Determine \( \mathcal{L}^{-1}\{F\}.**
\[ sF(s) - 3F(s) = \frac{s + 1}{s^2 + 6s + 9} \]
- **Click here to view the table of Laplace transforms.**
- **Click here to view the table of properties of Laplace transforms.**
\[ \mathcal{L}^{-1}\{F\} = \square \]
*Explanation:*
This equation involves using the inverse Laplace transform, denoted as \( \mathcal{L}^{-1} \), to find the original function \( F(t) \) in the time domain from its Laplace-transformed function. The expression given is \( sF(s) - 3F(s) = \frac{s + 1}{s^2 + 6s + 9} \).
To solve for \( F(t) \), you can utilize the tables of Laplace transforms and their properties linked in the document. The term \( \frac{s + 1}{s^2 + 6s + 9} \) suggests that partial fraction decomposition or known transform pairs might be useful for finding \( \mathcal{L}^{-1}\{F\} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F502330eb-9fa1-41eb-8716-374edceb5782%2Ffbb47a5a-582b-436c-9d76-bcc849b31d74%2Foegphnp_processed.png&w=3840&q=75)
Transcribed Image Text:**Determine \( \mathcal{L}^{-1}\{F\}.**
\[ sF(s) - 3F(s) = \frac{s + 1}{s^2 + 6s + 9} \]
- **Click here to view the table of Laplace transforms.**
- **Click here to view the table of properties of Laplace transforms.**
\[ \mathcal{L}^{-1}\{F\} = \square \]
*Explanation:*
This equation involves using the inverse Laplace transform, denoted as \( \mathcal{L}^{-1} \), to find the original function \( F(t) \) in the time domain from its Laplace-transformed function. The expression given is \( sF(s) - 3F(s) = \frac{s + 1}{s^2 + 6s + 9} \).
To solve for \( F(t) \), you can utilize the tables of Laplace transforms and their properties linked in the document. The term \( \frac{s + 1}{s^2 + 6s + 9} \) suggests that partial fraction decomposition or known transform pairs might be useful for finding \( \mathcal{L}^{-1}\{F\} \).
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