What is laplacetransform t^3*cos(3t) ?

The answer to laplacetransform t^3*cos(3t) is -(6(-s^4+54s^2-81))/((s^2+9)^4)

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laplacetransform t^3*cos(3t)

laplace transform

t^{3} \cdot cos (3 t)

- \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

Solution steps

L {t^{3} cos (3 t)}

Use Laplace Transform table:

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

For

t^{3} cos (3 t) : f (t) = cos (3 t), k = 3

= (- 1)^{3} \frac{d ^{3}}{d s ^{3}} (L {cos (3 t)})

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

\frac{d ^{3}}{d s ^{3}} (\frac{s}{s ^{2} + 9}) = \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

= (- 1)^{3} \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

= - \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

x \to 0 lim (x + 1)

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

\frac{d ^{6}}{d x ^{6}} (x^{- 7})

\int (3 x + cos (x)) d x

\int \frac{1}{x ^{2} - 6 x + 25} d x

What is laplacetransform t^3*cos(3t) ?
The answer to laplacetransform t^3*cos(3t) is -(6(-s^4+54s^2-81))/((s^2+9)^4)