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}

\int \frac{12 x - 2}{6 x ^{2} - 2 x} d x

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

\frac{d}{d x} (50 x - x^{2})

\frac{d y}{d x} = k (y - 70)

f (x) = 3 x 2 - 5 x, at x = - 1

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