What is laplacetransform t^2cos(-2t) ?

The answer to laplacetransform t^2cos(-2t) is -(2s(-s^2+12))/((s^2+4)^3)

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laplacetransform t^2cos(-2t)

laplace transform

t^{2} cos (- 2 t)

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

Solution steps

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

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

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

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

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

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

\int_{0}^{1} \frac{2}{1 + x ^{2}} d x

y^{^{'}} = \frac{1}{e ^{y} - x}

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

\frac{d}{d x} (∣ ln ∣ x ∣ ∣)

\frac{d}{d x} (\frac{5}{2})

What is laplacetransform t^2cos(-2t) ?
The answer to laplacetransform t^2cos(-2t) is -(2s(-s^2+12))/((s^2+4)^3)