What is inverse oflaplace (e^{-s})/(s^5) ?

The answer to inverse oflaplace (e^{-s})/(s^5) is (\H(t-1)(t-1)^4)/(24)

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inverse oflaplace (e^{-s})/(s^5)

inverse laplace

\frac{e ^{- s}}{s ^{5}}

\frac{H ( t - 1 ) ( t - 1 ) ^{4}}{24}

Solution steps

L^{- 1} {\frac{e ^{- s}}{s ^{5}}}

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 ^{5}}} (t - 1)

L^{- 1} {\frac{1}{s ^{5}}} : \frac{t ^{4}}{24}

= H (t - 1) \frac{( t - 1 ) ^{4}}{24}

= \frac{H ( t - 1 ) ( t - 1 ) ^{4}}{24}

x \to 0 lim (\frac{( 4 ^{x} - 1 )}{x})

\int sin (7 x - 10) d x

x \to \infty lim (4 x^{3})

\frac{d}{d x} (12 x^{3})

\frac{\partial}{\partial y} (ln (\frac{x ^{2}}{y}))

What is inverse oflaplace (e^{-s})/(s^5) ?
The answer to inverse oflaplace (e^{-s})/(s^5) is (\H(t-1)(t-1)^4)/(24)