What is laplacetransform te^{-t} ?

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

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laplacetransform te^{-t}

laplace transform

t 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}}

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

\frac{d}{d x} (\frac{2 x - 1}{x + 2})

y^{^{'}} = e^{8 x} - 6 x

f (x) = \frac{40 t}{t + 2}

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