To determine The value of f ( t ) using the theorem (3) if the value of L ( f ) = 3 s + 4 s 4 + k 2 s 2 . Explanation Formula used: 1. The Laplace transform of the function e a t is 1 s − a . 2. The Laplace transform of the function sin a t is a s 2 + a 2 . 3. The Laplace transform of the function cos a t is s s 2 + a 2 Theorem used: Laplace transform of integral (Theorem 3) Let F ( s ) denote the transform of a function f ( t ) which is piecewise continuous for t ≥ 0 and satisfies a growth restriction Then, for s > 0 , s > k , and t > 0 , L ( ∫ 0 t f ( τ ) d τ ) = 1 s F ( s ) , thus ∫ 0 t f ( τ ) d τ = L − 1 ( 1 s F ( s ) ) ( 1 ) . Calculation: It is given that F ( s ) = 3 s + 4 s 4 + k 2 s 2 . Rewrite the given transform of the function as follows. F ( s ) = 3 s + 4 s 4 + k 2 s 2 = 3 s + 4 s 2 ( s 2 + k 2 ) = 1 s 2 ( 3 s s 2 + k 2 + 4 k ⋅ k s 2 + k 2 ) Let f ( t ) be the inverse transform of F ( s ) . Now, using the above mentioned theorem observe that the inverse transform of F ( s ) can be obtained using the formula f ( t ) = ∫ 0 t g ( τ ) d τ where g ( t ) = L − 1 ( 3 s s 2 + k 2 + 4 k ⋅ k s 2 + k 2 ) . Obtain the value of g ( t ) = L − 1 ( 3 s s 2 + k 2 + 4 k ⋅ k s 2 + k 2 ) as shown below

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Author: Kreyszig

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Chapter 6.2, Problem 28P

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The value of f(t) using the theorem (3) if the value of L(f)=3s+4s4+k2s2.

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Chapter 6.2, Problem 27P

Chapter 6.2, Problem 29P

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