What is laplacetransform t^3sin(2t) ?

The answer to laplacetransform t^3sin(2t) is -(48s(-s^2+4))/((s^2+4)^4)

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laplacetransform t^3sin(2t)

laplace transform

t^{3} sin (2 t)

- \frac{48 s ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{4}}

Solution steps

L {t^{3} 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^{3} sin (2 t) : f (t) = sin (2 t), k = 3

= (- 1)^{3} \frac{d ^{3}}{d s ^{3}} (L {sin (2 t)})

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

\frac{d ^{3}}{d s ^{3}} (\frac{2}{s ^{2} + 4}) = \frac{48 s ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{4}}

= (- 1)^{3} \frac{48 s ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{4}}

= - \frac{48 s ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{4}}

x \cdot \frac{d y}{d x} - y = x^{2} sin (x)

\int y - 1 d y

\frac{3}{( s + 2 ) ^{2} + 9}

\int \frac{1}{9 sin ^{2} ( θ )} d θ

\int_{0}^{π} sin^{5} (\frac{x}{2}) d x

What is laplacetransform t^3sin(2t) ?
The answer to laplacetransform t^3sin(2t) is -(48s(-s^2+4))/((s^2+4)^4)