Question

The Laplace inverse of $\frac{s}{s^2 - 1}$ is, i.e., $\mathcal{L}^{-1} \left( \frac{s}{s^2 - 1} \right)$ a) cosh t b) sinh t c) cost d) sint

          The Laplace inverse of $\frac{s}{s^2 - 1}$ is, i.e., $\mathcal{L}^{-1} \left( \frac{s}{s^2 - 1} \right)$ 
a) cosh t
b) sinh t
c) cost
d) sint

The Laplace inverse of (s)/(s^2 - 1) is, i.e., ℒ^-1( (s)/(s^2 - 1))
a) cosh t
b) sinh t
c) cost
d) sint

Added by Shelley F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

04/24/2023

Step 1

The denominator is s^2 - 1, which can be factored as (s-1)(s+1). Show more…

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The Laplace inverse of (s)/(s^(2)-1) is, i.e., L^(-1)((s)/(s^(2)-1)) a) cosht b) sinht c) cost d) sint a cosh t bsinh t c cost d sint

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Question

The Laplace inverse of $\frac{s}{s^2 - 1}$ is, i.e., $\mathcal{L}^{-1} \left( \frac{s}{s^2 - 1} \right)$ a) cosh t b) sinh t c) cost d) sint

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