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}}

f (x) = x + 7 x

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

\frac{\partial}{\partial x} ((x + y)^{5} + (x - y)^{5})

f (x) = tan^{2} (x), at x = - \frac{π}{4}

f (x) = 13 x e^{x}

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