Question

Find the inverse Laplace transform $f(t) = \mathcal{L}^{-1} \{F(s)\}$ of the function $F(s) = \frac{9}{s^2 + 64} + \frac{3s}{s^2 + 16}$ $f(t) = \mathcal{L}^{-1} \left\{ \frac{9}{s^2 + 64} + \frac{3s}{s^2 + 16} \right\} = 0.125 \sin(8t) + 3 \cos(4t)$

          Find the inverse Laplace transform $f(t) = \mathcal{L}^{-1} \{F(s)\}$ of the function $F(s) = \frac{9}{s^2 + 64} + \frac{3s}{s^2 + 16}$
$f(t) = \mathcal{L}^{-1} \left\{ \frac{9}{s^2 + 64} + \frac{3s}{s^2 + 16} \right\} = 0.125 \sin(8t) + 3 \cos(4t)$

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$Find the inverse Laplace transform f(t) = ℒ^-1{F(s)} of the function F(s) = (9)/(s^2 + 64) + (3s)/(s^2 + 16) f(t) = ℒ^-1{(9)/(s^2 + 64) + (3s)/(s^2 + 16)} = 0.125 sin(8t) + 3 cos(4t)$

Added by Mark G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Teresa Fuston

12/28/2021

Step 1

Step 1: The given expression is 9(3s + 16) / (s^2 + 64) + f = c. Show more…

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9 3s $2 + 16 s2+64 f=c 9 3s +16 0.125sin8r+3 cos4r help.(formulas) s264

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