Question

Find the inverse Laplace transform $f(t) = \mathcal{L}^{-1} \{F(s)\}$ of the function $F(s) = \frac{8s - 28}{s^2 - 6s + 13}$ $f(t) = \mathcal{L}^{-1} \left\{ \frac{8s - 28}{s^2 - 6s + 13} \right\} = $ help (formulas)

          Find the inverse Laplace transform $f(t) = \mathcal{L}^{-1} \{F(s)\}$ of the function
$F(s) = \frac{8s - 28}{s^2 - 6s + 13}$
$f(t) = \mathcal{L}^{-1} \left\{ \frac{8s - 28}{s^2 - 6s + 13} \right\} = $ help (formulas)

$Find the inverse Laplace transform f(t) = ℒ^-1{F(s)} of the function F(s) = (8s - 28)/(s^2 - 6s + 13) f(t) = ℒ^-1{(8s - 28)/(s^2 - 6s + 13)} = help (formulas)$

Added by Monica D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/18/2023

Step 1

The denominator s^2 - 6s + 13 can be factored as (s - 3)^2 + 4. Show more…

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Find the inverse Laplace transform f(t)=L^(-1){F(s)} of the function F(s)=(8s-28)/(s^(2)-6s+13). f(t)=L^(-1){(8s-28)/(s^(2)-6s+13)}=, help (formulas) Find the inverse Laplace transform f(t) = -1{F(s)} of the function 8s - 28 F s2-6s+13 8s - 28 help (formulas) s2-6s+13