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^{2} + y^{2} = 2

\frac{\partial}{\partial x} (3 x (4 + y)^{- 1})

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

\frac{\partial}{\partial x} (e^{3 y})

\int \frac{z}{z ^{4} + 4} d z

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)