Determine the inverse Laplace transform of the function below. 4s +37 s² +6s+34 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. L-1 4s +37 $²2+6s+34 Properties of Laplace Transforms L{f+g} = L{f} + L{g} L{cf} = c£{f} for any constant c Leatf(t)} (s) = L{f}(s-a) L {f'} (s) = s£{f}(s) - f(0) L {f'' (s) = s² L{f}(s) - sf(0) - f'(0) L {f(n)} (s) = s£{f}(s)-s-1f(0)-s-2f'(0)-... -f(n-1) (0) dn L {t^f(t)} (s) = (-1)^- dsn = £¯¹ {F₁} + £¯ £¯1{F₁+F2} L1¹{CF} = c£¯¹{F} (L{f}(s)) Table of Laplace Transforms X f(t) 1 eat t", n=1,2,... sin bt cos bt eat,n=1,2,.... eat sin bt eat cos bt F(s) = L{f}(s) 1 s 1 s-a n! Sn+18>0 ,S>0 ,S> 0 b s² + b² ,s > 0 S s²+ b² n! (s-a)n +1 b ,S>0 .s> a (s-a)² + b² s-a (s-a)² + b² .s> a .s> a
Determine the inverse Laplace transform of the function below. 4s +37 s² +6s+34 Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. L-1 4s +37 $²2+6s+34 Properties of Laplace Transforms L{f+g} = L{f} + L{g} L{cf} = c£{f} for any constant c Leatf(t)} (s) = L{f}(s-a) L {f'} (s) = s£{f}(s) - f(0) L {f'' (s) = s² L{f}(s) - sf(0) - f'(0) L {f(n)} (s) = s£{f}(s)-s-1f(0)-s-2f'(0)-... -f(n-1) (0) dn L {t^f(t)} (s) = (-1)^- dsn = £¯¹ {F₁} + £¯ £¯1{F₁+F2} L1¹{CF} = c£¯¹{F} (L{f}(s)) Table of Laplace Transforms X f(t) 1 eat t", n=1,2,... sin bt cos bt eat,n=1,2,.... eat sin bt eat cos bt F(s) = L{f}(s) 1 s 1 s-a n! Sn+18>0 ,S>0 ,S> 0 b s² + b² ,s > 0 S s²+ b² n! (s-a)n +1 b ,S>0 .s> a (s-a)² + b² s-a (s-a)² + b² .s> a .s> a
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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Question
![**Determine the Inverse Laplace Transform**
**Problem Statement:**
Determine the inverse Laplace transform of the function below.
\[ \frac{4s + 37}{s^2 + 6s + 34} \]
[Click here to view the table of Laplace transforms.](#)
[Click here to view the table of properties of Laplace transforms.](#)
**Inverse Laplace Transform Expression:**
\[ \mathcal{L}^{-1} \left\{ \frac{4s + 37}{s^2 + 6s + 34} \right\} = \boxed{\phantom{placeholder}} \]
---
**Properties of Laplace Transforms**
- \(\mathcal{L}\{f + g\} = \mathcal{L}\{f\} + \mathcal{L}\{g\}\)
- \(\mathcal{L}\{cf\} = c\mathcal{L}\{f\} \) for any constant \(c\)
- \(\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s-a)\)
- \(\mathcal{L}\{f'(t)\}(s) = s\mathcal{L}\{f(t)\} - f(0)\)
- \(\mathcal{L}\{f''(t)\}(s) = s^2\mathcal{L}\{f(t)\} - sf(0) - f'(0)\)
- \(\mathcal{L}\{f^{(n)}(t)\}(s) = s^n\mathcal{L}\{f(t)\}(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)\)
- \(\mathcal{L}\{f^{(n)}(t)\}(s) = (-1)^n \frac{d^n}{ds^n}(\mathcal{L}\{f(t)\})\)
- \(\mathcal{L}^{-1}\{F_1 + F_2\} = \mathcal{L}^{-1}\{F_1\} + \mathcal{L}^{-1}\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F047a7e6a-f025-4b5b-ab83-4ffe14f69253%2F88253b07-3769-4196-9dcd-89bd80b3db0c%2F0n5ulf_processed.png&w=3840&q=75)
Transcribed Image Text:**Determine the Inverse Laplace Transform**
**Problem Statement:**
Determine the inverse Laplace transform of the function below.
\[ \frac{4s + 37}{s^2 + 6s + 34} \]
[Click here to view the table of Laplace transforms.](#)
[Click here to view the table of properties of Laplace transforms.](#)
**Inverse Laplace Transform Expression:**
\[ \mathcal{L}^{-1} \left\{ \frac{4s + 37}{s^2 + 6s + 34} \right\} = \boxed{\phantom{placeholder}} \]
---
**Properties of Laplace Transforms**
- \(\mathcal{L}\{f + g\} = \mathcal{L}\{f\} + \mathcal{L}\{g\}\)
- \(\mathcal{L}\{cf\} = c\mathcal{L}\{f\} \) for any constant \(c\)
- \(\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s-a)\)
- \(\mathcal{L}\{f'(t)\}(s) = s\mathcal{L}\{f(t)\} - f(0)\)
- \(\mathcal{L}\{f''(t)\}(s) = s^2\mathcal{L}\{f(t)\} - sf(0) - f'(0)\)
- \(\mathcal{L}\{f^{(n)}(t)\}(s) = s^n\mathcal{L}\{f(t)\}(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)\)
- \(\mathcal{L}\{f^{(n)}(t)\}(s) = (-1)^n \frac{d^n}{ds^n}(\mathcal{L}\{f(t)\})\)
- \(\mathcal{L}^{-1}\{F_1 + F_2\} = \mathcal{L}^{-1}\{F_1\} + \mathcal{L}^{-1}\
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