What is inverse oflaplace (2se^{-3s})/(3s^2+27) ?

The answer to inverse oflaplace (2se^{-3s})/(3s^2+27) is \H(t-3) 2/3 cos(3(t-3))

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inverse oflaplace (2se^{-3s})/(3s^2+27)

inverse laplace

\frac{2 s e ^{- 3 s}}{3 s ^{2} + 27}

H (t - 3) \frac{2}{3} cos (3 (t - 3))

Solution steps

L^{- 1} {\frac{2 e ^{- 3 s} s}{3 s ^{2} + 27}}

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 - 3) L^{- 1} {\frac{2 s}{3 s ^{2} + 27}} (t - 3)

L^{- 1} {\frac{2 s}{3 s ^{2} + 27}} : \frac{2}{3} cos (3 t)

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

\frac{d}{d x} (\frac{1}{1 - x ^{2} - y})

\int \frac{e ^{\frac{4}{t}}}{t ^{2}} d t

\frac{d}{d x} (\frac{8 - x e ^{x}}{x + e ^{x}})

\int + \frac{1}{- x ^{2} + 8 x - 15} d x

(\frac{x ^{2}}{e ^{2 x}})^{6}

What is inverse oflaplace (2se^{-3s})/(3s^2+27) ?
The answer to inverse oflaplace (2se^{-3s})/(3s^2+27) is \H(t-3) 2/3 cos(3(t-3))