Give the inverse Laplace transform of -55 e -Ss 2 F(s) = 4 e 5: as a function of x. Note: The function u below is the unit step function, which is also known as the Heaviside function. a) f(x) =4 u(x – 5) x – 2– 3 4(x – 5) b) O (x) =-4(x - 5) x – 2+4(x – 5) c) 1х) -3и (х — 5) х— 2- и (х — 5) d) O f(x) =u(x - 5) x – 2+ 5u(x – 5) (x) =u(x – 5) x - 2- 9 4(x – 5) O 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
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**Problem Statement:**

Give the inverse Laplace transform of 

\[ 
F(s) = -\frac{2}{s} + \frac{e^{-5s}}{s^2} - \frac{4e^{-5s}}{s}
\]

as a function of \( x \).

**Note:** The function \( u \) below is the unit step function, which is also known as the Heaviside function.

**Options:**

a) \( f(x) = 4u(x-5)x - 2 - 3u(x-5) \)

b) \( f(x) = -u(x-5)x - 2 + u(x-5) \)

c) \( f(x) = 3u(x-5)x - 2 - u(x-5) \)

d) \( f(x) = u(x-5)x - 2 + 5u(x-5) \)

e) \( f(x) = u(x-5)x - 2 - 9u(x-5) \)

f) None of the above.
Transcribed Image Text:**Problem Statement:** Give the inverse Laplace transform of \[ F(s) = -\frac{2}{s} + \frac{e^{-5s}}{s^2} - \frac{4e^{-5s}}{s} \] as a function of \( x \). **Note:** The function \( u \) below is the unit step function, which is also known as the Heaviside function. **Options:** a) \( f(x) = 4u(x-5)x - 2 - 3u(x-5) \) b) \( f(x) = -u(x-5)x - 2 + u(x-5) \) c) \( f(x) = 3u(x-5)x - 2 - u(x-5) \) d) \( f(x) = u(x-5)x - 2 + 5u(x-5) \) e) \( f(x) = u(x-5)x - 2 - 9u(x-5) \) f) None of the above.
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