What is laplacetransform te^{-t}sin(2t) ?

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

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laplacetransform te^{-t}sin(2t)

laplace transform

t e^{- t} sin (2 t)

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

Solution steps

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

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

L {e^{- t} sin (2 t)} : \frac{2}{( s + 1 ) ^{2} + 4}

\frac{d}{d s} (\frac{2}{( s + 1 ) ^{2} + 4}) = - \frac{2 ( 2 s + 2 )}{( s ^{2} + 2 s + 5 ) ^{2}}

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

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

y^{^{'}} - \frac{4}{t} y = t^{5} e^{t}

x \to 1 - lim (\frac{x - 3}{x - 1})

\int y (2 y + 1) e^{4 y^{3} + 3 y^{2} - 4} d y

5 x^{2} - x^{3}

\int (5 - \frac{2}{3} x^{2} + \frac{3}{4} x^{3}) d x

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