Explanation The given function is e − π s s 2 + 1 . From inverse form of Second Translation theorem, if f ( t ) = L − 1 { F ( s ) } and a > 0 , then L − 1 { e − a s F ( s ) } = f ( t − a ) U ( t − a ) . Here, F ( s ) = 1 s 2 + 1 and a = π

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 4.3, Problem 45E

To determine

The inverse Laplace transformation of the given function.

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Chapter 4.3, Problem 44E

Chapter 4.3, Problem 46E

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For f(x) and g(x) given in Problems 35–38, find (a) (f + g)(x) (b) (f – g)(x) (c) (f'g)(x) (d) (f/g)(x) 35. f(x) = 3x g(x) = x' 36. f(x) = Vx g(x) = 1/x 37. f(x) = V2x g(x) = x² 38. f(x) = (x – 1)? g(x) = 1 – 2x Click to %3D For f(x) and g(x) given in Problems 39–42, find (a) (fº g)(x) (b) (g •f)(x) (c) ƒ(f(x)) (d) f(x) = (f·f)(x) 39. f(x) = (x – 1)³ g(x) = 1 – 2x 40. f(x) = 3x g(x) = x' – 1 41. f(x) = 2Vx g(x) = x* + 5 %3D %3D 1 42. f(x) = g(x) = 4x + 1

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Chapter 4 Solutions

Advanced Engineering Mathematics

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The inverse Laplace transformation of the given function.

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Chapter 4 Solutions