What is laplacetransform te^{-5t} ?

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

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laplacetransform te^{-5t}

laplace transform

t 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}}

5 t ln (t)

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

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

\frac{d}{d x} (co tan (x) - tan (x))

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