What is laplacetransform t*e^{-5t} ?

The answer to laplacetransform t*e^{-5t} is 1/((s+5)^2)

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laplacetransform t*e^{-5t}

laplace transform

t \cdot e^{- 5 t}

\frac{1}{( s + 5 ) ^{2}}

Solution steps

L {e^{- 5 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^{- 5 t} : f (t) = e^{- 5 t}, k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 5 t}})

L {e^{- 5 t}} : \frac{1}{s + 5}

\frac{d}{d s} (\frac{1}{s + 5}) = - \frac{1}{( s + 5 ) ^{2}}

= (- 1)^{1} (- \frac{1}{( s + 5 ) ^{2}})

= \frac{1}{( s + 5 ) ^{2}}

f (x) = 5 x + 4

x \to - 7 lim (\frac{1}{x ^{2} ( x + 7 )})

(81 - x^{2}) y^{^{'}} = 7 y

\frac{d}{d x} (\frac{8}{e ^{x} + e ^{- x}})

\int \frac{e ^{3 x}}{4 - 4 e ^{6 x}} d x

What is laplacetransform t*e^{-5t} ?
The answer to laplacetransform t*e^{-5t} is 1/((s+5)^2)