What is laplacetransform te^{-2t} ?

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

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laplacetransform te^{-2t}

laplace transform

t e^{- 2 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

t e^{- 2 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}}

\frac{\partial}{\partial x} (\frac{y}{x + 1})

\frac{s}{1}

\frac{\partial}{\partial x} (- x^{2} y + \frac{1}{y})

\int_{1}^{3} 2^{x} d x

f (x) = x^{2} + 36

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