F(s) = s (s + 2) 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) = -2 e 4* + u(x – 2) – u(x – 2) e2*+4 -2x b) O f(x) =5 e2* – u (x – 2) +u(x – 2) e-2* +4 c) f(x) =4 e* -2x – 2 u(x – 2) + 2u(x - 2) e -2x +4 a) O ste) --362*+ u(x– 2) – u(x – 2) e*2** -2x d) O f(x) =-3 e -2x -2x +4 (x) = -2 e * – 2 u(x – 2) + 2u(x – 2) e ** f) O None of the above.
F(s) = s (s + 2) 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) = -2 e 4* + u(x – 2) – u(x – 2) e2*+4 -2x b) O f(x) =5 e2* – u (x – 2) +u(x – 2) e-2* +4 c) f(x) =4 e* -2x – 2 u(x – 2) + 2u(x - 2) e -2x +4 a) O ste) --362*+ u(x– 2) – u(x – 2) e*2** -2x d) O f(x) =-3 e -2x -2x +4 (x) = -2 e * – 2 u(x – 2) + 2u(x – 2) e ** 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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Transcribed Image Text:Give the inverse Laplace transform of
-35+5e 25
F(s) =
5 (5 + 2)
as a function of x.
Note: The function u below is the unit step function, which is also known as the Heaviside function.
-2x
f(x) =-2 e * +u(x - 2) – u(x – 2) e
e2x +4
-2x
b)
(x) =5 e 4* – u(x - 2) + u(x– 2) e 2x +4
-2x
-2x +4
c) O f(x) =4 e4* – 2 u(x – 2) + 2 u(x – 2) e
5
d) O sx) = -3e 2* +u(x – 2) – u(z – 2) e 2**
-2x
-2x +4
-2x
f(x) = -2 e 2* – 2 u(x- 2) + 2 u(x– 2) e * *4
f) O None of the above.
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