What is inverse oflaplace (e^{-2s})/(s^4) ?

The answer to inverse oflaplace (e^{-2s})/(s^4) is (\H(t-2)(t-2)^3)/6

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inverse oflaplace (e^{-2s})/(s^4)

inverse laplace

\frac{e ^{- 2 s}}{s ^{4}}

\frac{H ( t - 2 ) ( t - 2 ) ^{3}}{6}

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{s ^{4}}}

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 - 2) L^{- 1} {\frac{1}{s ^{4}}} (t - 2)

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= H (t - 2) \frac{( t - 2 ) ^{3}}{6}

= \frac{H ( t - 2 ) ( t - 2 ) ^{3}}{6}

What is inverse oflaplace (e^{-2s})/(s^4) ?
The answer to inverse oflaplace (e^{-2s})/(s^4) is (\H(t-2)(t-2)^3)/6