What is inverse oflaplace e^{-s}(1/(s^2)) ?

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

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inverse oflaplace e^{-s}(1/(s^2))

inverse laplace

e^{- s} (\frac{1}{s ^{2}})

H (t - 1) (t - 1)

Solution steps

L^{- 1} {e^{- s} (\frac{1}{s ^{2}})}

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

L^{- 1} {\frac{1}{s ^{2}}} : t

= H (t - 1) (t - 1)

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