What is laplacetransform t^2cos(kt) ?

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

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laplacetransform t^2cos(kt)

laplace transform

t^{2} cos (k t)

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

Solution steps

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

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

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

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

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

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

\int_{1}^{2} (3 x^{2} + 2 x + 1) d x

\frac{\partial}{\partial y} ((x y)^{\frac{1}{3}})

\int x 5 - 2 x d x

\int_{0}^{7} x e^{- 4 x} d x

\frac{d}{d x} (a x^{3})

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