What is laplacetransform t^2cos(8t) ?

The answer to laplacetransform t^2cos(8t) is -(2s(-s^2+192))/((s^2+64)^3)

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laplacetransform t^2cos(8t)

laplace transform

t^{2} cos (8 t)

- \frac{2 s ( - s ^{2} + 192 )}{( s ^{2} + 64 ) ^{3}}

Solution steps

L {t^{2} cos (8 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 (8 t) : f (t) = cos (8 t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {cos (8 t)})

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

\frac{d ^{2}}{d s ^{2}} (\frac{s}{s ^{2} + 64}) = - \frac{2 s ( - s ^{2} + 192 )}{( s ^{2} + 64 ) ^{3}}

= (- 1)^{2} (- \frac{2 s ( - s ^{2} + 192 )}{( s ^{2} + 64 ) ^{3}})

= - \frac{2 s ( - s ^{2} + 192 )}{( s ^{2} + 64 ) ^{3}}

\frac{d}{d x} (x^{3} - 8 x^{2} + 31 x)

\frac{d ^{2}}{d x ^{2}} (x^{2} e^{x})

n = 0 \sum \infty - 3^{n}

\frac{7}{x}

\int \frac{1}{- x ^{2} + 4 x - 1} d x

What is laplacetransform t^2cos(8t) ?
The answer to laplacetransform t^2cos(8t) is -(2s(-s^2+192))/((s^2+64)^3)