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Question: 8) Find the inverse Laplace transform F(s) = cot^-1 (s – 2)

8) Find the inverse Laplace transform

F(s) = cot^-1 (s – 2)

Answer

F(s) = cot^-1(s-2)

Because we all know this already

f(t) = L-1 [ F(s) ] = ( -1/ t) L-1[F'(s) ]

so we first calculate F'(s) ,

F'(s) = -1 / [(s-2)^2 + 1 ]

Now, put this value

f(t) = (-1/t) L-1[  -1 / [(s-2)^2 + 1 ] ]

= (1/t) L-1[1 / (s-2)^2 + 1]

By using the inverse laplace formula,

(e^at)sinbt = b /(s-a)^2 + b*b

= (1/t )*[(e^2t)sint ]

= (e^2t)sint / t

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