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

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

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laplacetransform e^{-2t}t

laplace transform

e^{- 2 t} t

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

Solution steps

L {e^{- 2 t} t}

Use Laplace Transform table:

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

For

e^{- 2 t} t : f (t) = e^{- 2 t}, k = 1

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

L {e^{- 2 t}} : \frac{1}{s + 2}

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

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

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

e^{x}, - 4

\frac{\partial}{\partial y} ((2 x + 3 y)^{10})

\frac{1 - 3 t}{4 + t}

\int \frac{1}{7 x + 1} d x

y = \frac{2}{x}, (4, \frac{1}{2})

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