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

The answer to inverse oflaplace (e^{-s})/(s^3) is (\H(t-1)(t-1)^2)/2

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

inverse laplace

\frac{e ^{- s}}{s ^{3}}

\frac{H ( t - 1 ) ( t - 1 ) ^{2}}{2}

Solution steps

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

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 ^{3}}} (t - 1)

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

= H (t - 1) \frac{( t - 1 ) ^{2}}{2}

= \frac{H ( t - 1 ) ( t - 1 ) ^{2}}{2}

\int \frac{1}{x + 1 + x ^{2}} d x

\frac{1}{s ^{3}}

x^{2} y^{^{'}} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

\int \frac{1}{e ^{x} + e ^{2 x}} d x

\int \frac{tan ^{3} ( ln ( x ) )}{7 x} d x

What is inverse oflaplace (e^{-s})/(s^3) ?
The answer to inverse oflaplace (e^{-s})/(s^3) is (\H(t-1)(t-1)^2)/2