What is inverse oflaplace (2e^{-2s})/(3s^2+12) ?

The answer to inverse oflaplace (2e^{-2s})/(3s^2+12) is \H(t-2) 1/3 sin(2(t-2))

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inverse oflaplace (2e^{-2s})/(3s^2+12)

inverse laplace

\frac{2 e ^{- 2 s}}{3 s ^{2} + 12}

H (t - 2) \frac{1}{3} sin (2 (t - 2))

Solution steps

L^{- 1} {\frac{2 e ^{- 2 s}}{3 s ^{2} + 12}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 2) L^{- 1} {\frac{2}{3 s ^{2} + 12}} (t - 2)

L^{- 1} {\frac{2}{3 s ^{2} + 12}} : \frac{1}{3} sin (2 t)

= H (t - 2) \frac{1}{3} sin (2 (t - 2))

What is inverse oflaplace (2e^{-2s})/(3s^2+12) ?
The answer to inverse oflaplace (2e^{-2s})/(3s^2+12) is \H(t-2) 1/3 sin(2(t-2))