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

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

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

inverse laplace

\frac{2}{s ^{2}} \cdot e^{- s}

H (t - 1) \cdot 2 (t - 1)

Solution steps

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

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

= H (t - 1) \cdot 2 (t - 1)

f^{^{'}} (x) = \frac{3}{4} \cdot (1 - x^{2})

\frac{d}{d x} (x^{2} + 6 x + 3)

\frac{d}{d x} (3 x^{2} - 12 x + 9)

11 t^{2} y^{^{''}} - 22 y = 9 t^{2} - 9

x \to 0 lim (cos (2 - 2 e^{x}))

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