What is inverse oflaplace 2/((s+1)^3) ?

The answer to inverse oflaplace 2/((s+1)^3) is e^{-t}t^2

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inverse oflaplace 2/((s+1)^3)

inverse laplace

\frac{2}{( s + 1 ) ^{3}}

e^{- t} t^{2}

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{2}{( s + 1 ) ^{3}} : a = - 1, F (s) = \frac{2}{s ^{3}}

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

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

= e^{- t} t^{2}

\frac{d}{d x} (x e^{\frac{y}{x}})

x \to 1 lim ((1 - x)^{2} + 2)

\frac{d}{d x} (\frac{2160}{x ^{2}})

x \to \infty lim (\frac{4}{0})

\int_{0}^{4} x^{2} (0.5 - 0.125 x) d x

What is inverse oflaplace 2/((s+1)^3) ?
The answer to inverse oflaplace 2/((s+1)^3) is e^{-t}t^2