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

The answer to inverse oflaplace (e^{-pis})/(s^3+s) is \H(t-pi)(1-cos(t-pi))

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

inverse laplace

\frac{e ^{- π s}}{s ^{3} + s}

H (t - π) (1 - cos (t - π))

Solution steps

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

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

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

= H (t - π) (H (t - π) - cos (t - π))

H (t - π) = {01 t < π t \geq π

= {0 1 - cos (t - π) t < π t \geq π

= H (t - π) (1 - cos (t - π))

y^{^{'}} = 0.5 x - 450

x \to - 1 lim (\frac{2 - 5 x}{2 + 2 x})

\frac{d}{d x} (3 x^{5} ln (x))

f (x) = \frac{4 x ^{2}}{x + 2}

3 + 6 cos (3 t)

What is inverse oflaplace (e^{-pis})/(s^3+s) ?
The answer to inverse oflaplace (e^{-pis})/(s^3+s) is \H(t-pi)(1-cos(t-pi))