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

The answer to laplacetransform t^2e^{-2t}sin(2t) is -(4(-3s^2-12s-8))/((s^2+4s+8)^3)

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

laplace transform

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

- \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

Solution steps

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

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{2}{( s + 2 ) ^{2} + 4}) = - \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

= (- 1)^{2} (- \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}})

= - \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

\frac{\partial}{\partial x} (2 x + y)

7 y^{^{'}} - 14 y = 0

\int tan (x) d x

\frac{- 3}{2 x}

What is laplacetransform t^2e^{-2t}sin(2t) ?
The answer to laplacetransform t^2e^{-2t}sin(2t) is -(4(-3s^2-12s-8))/((s^2+4s+8)^3)