What is inverse oflaplace (e^{-2s})/(s^3) ?

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

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inverse oflaplace (e^{-2s})/(s^3)

inverse laplace

\frac{e ^{- 2 s}}{s ^{3}}

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

Solution steps

L^{- 1} {\frac{e ^{- 2 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 - 2) L^{- 1} {\frac{1}{s ^{3}}} (t - 2)

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

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

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

\int x \cdot e^{x} d x

2 sin (t) cos (t)

y^{^{''}} - 4 y^{^{'}} + 5 = 0

f (x) = \frac{5}{2 5 x + 6}, at a = 2

y = \frac{x ^{3}}{9}, 0 \leq x \leq 2

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