Give the inverse Laplace transform of -25 + (s – 4) e +9 F(s) = as a function of x. Note: The function u below is the unit step function, which is also known as the Heaviside function. a) O f(x) = -2 cos(3 x) – 4(x – T) cos( 3x) + 4 u(x- 7) sin(3x) b) O f(x) = -2 cos( 3 x) + u(x - a) cos( 3 x) + 5 -(x- x) sin(3x) 3 f(x) =2 sin(3x) + 3u(x- n) cos(3 x) + 2u(x – n) sin(3x) f(x) =2 sin(3x) – 3 u(x – 1) cos(3x) + 5 (x-x) sin(3x) 5 4 (x – T) sin (3x) -3x f(x) = -2 e3*+ u(x – n) cos(3 x) + 3 f) O None of the above.
Give the inverse Laplace transform of -25 + (s – 4) e +9 F(s) = as a function of x. Note: The function u below is the unit step function, which is also known as the Heaviside function. a) O f(x) = -2 cos(3 x) – 4(x – T) cos( 3x) + 4 u(x- 7) sin(3x) b) O f(x) = -2 cos( 3 x) + u(x - a) cos( 3 x) + 5 -(x- x) sin(3x) 3 f(x) =2 sin(3x) + 3u(x- n) cos(3 x) + 2u(x – n) sin(3x) f(x) =2 sin(3x) – 3 u(x – 1) cos(3x) + 5 (x-x) sin(3x) 5 4 (x – T) sin (3x) -3x f(x) = -2 e3*+ u(x – n) cos(3 x) + 3 f) O None of the above.
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**
**Objective:** Determine the inverse Laplace transform of the given function \( F(s) \) as a function of \( x \).
**Function:**
\[
F(s) = \frac{-2s + (s - 4) e^{-πs}}{s^2 + 9}
\]
**Note:** The function \( u \) below is the unit step function, which is also known as the Heaviside function.
**Options:**
a) \( f(x) = -2 \cos(3x) - u(x - \pi) \cos(3x) + \frac{4}{3} u(x - \pi) \sin(3x) \)
b) \( f(x) = -2 \cos(3x) + u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
c) \( f(x) = 2 \sin(3x) + 3 u(x - \pi) \cos(3x) + 2 u(x - \pi) \sin(3x) \)
d) \( f(x) = 2 \sin(3x) - 3 u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
e) \( f(x) = -2 e^{-3x} + 4 u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
f) \( \) None of the above.
The problem involves using the inverse Laplace transform to convert a function from the \( s \)-domain back into the time domain, \( x \), using potential expressions involving trigonometric functions and the unit step function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88afae02-80d5-49da-ad82-a7933e6f4445%2F8250103f-cd4e-4f19-8ae0-47f58bab8729%2Fzqdjaib_processed.png&w=3840&q=75)
Transcribed Image Text:**Inverse Laplace Transform Problem**
**Objective:** Determine the inverse Laplace transform of the given function \( F(s) \) as a function of \( x \).
**Function:**
\[
F(s) = \frac{-2s + (s - 4) e^{-πs}}{s^2 + 9}
\]
**Note:** The function \( u \) below is the unit step function, which is also known as the Heaviside function.
**Options:**
a) \( f(x) = -2 \cos(3x) - u(x - \pi) \cos(3x) + \frac{4}{3} u(x - \pi) \sin(3x) \)
b) \( f(x) = -2 \cos(3x) + u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
c) \( f(x) = 2 \sin(3x) + 3 u(x - \pi) \cos(3x) + 2 u(x - \pi) \sin(3x) \)
d) \( f(x) = 2 \sin(3x) - 3 u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
e) \( f(x) = -2 e^{-3x} + 4 u(x - \pi) \cos(3x) + \frac{5}{3} u(x - \pi) \sin(3x) \)
f) \( \) None of the above.
The problem involves using the inverse Laplace transform to convert a function from the \( s \)-domain back into the time domain, \( x \), using potential expressions involving trigonometric functions and the unit step function.
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