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

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

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inverse oflaplace (e^{-pi(s)})/(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))

\frac{d}{d x} ((x + \frac{1}{x})^{5})

\frac{d}{d x} ((x)^{x})

\int_{0}^{\infty} u e^{- u^{2}} d u

\frac{s}{s ^{2} + 10 s + 9.8}

\frac{d y}{d x} = 2 \cdot \frac{y}{x} + (\frac{y}{x})^{2}

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