What is laplacetransform e^{-t}t^2 ?

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

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laplacetransform e^{-t}t^2

laplace transform

e^{- t} t^{2}

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

Solution steps

L {e^{- t} t^{2}}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

e^{- t} t^{2} : f (t) = e^{- t}, k = 2

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

L {e^{- t}} : \frac{1}{s + 1}

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

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

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

\int 2 x 2 - x^{2} d x

\frac{d}{d x} (\frac{x ^{3} + x}{x ^{4} - 2})

y = 2 e^{- x} + e^{3 x}

n = 1 \sum \infty \frac{n}{4 n ^{3} + 1}

h (x) = sec^{6} (x)

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