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

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

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laplacetransform t*e^{-t}

laplace transform

t \cdot e^{- t}

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

Solution steps

L {e^{- t} t}

Use Laplace Transform table:

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

For

t e^{- t} : f (t) = e^{- t}, k = 1

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

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

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

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

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

f (x) = x^{3} e^{x}

\int \frac{1}{2 x ^{2} - 4 x + 12} d x

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

\int y e^{- y^{2}} d y

y = - 3 - 7 x^{2}

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