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

The answer to laplacetransform t^3*sin(3t) is -(72s(-s^2+9))/((s^2+9)^4)

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laplacetransform t^3*sin(3t)

laplace transform

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

- \frac{72 s ( - s ^{2} + 9 )}{( s ^{2} + 9 ) ^{4}}

Solution steps

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

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

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

\frac{d ^{3}}{d s ^{3}} (\frac{3}{s ^{2} + 9}) = \frac{72 s ( - s ^{2} + 9 )}{( s ^{2} + 9 ) ^{4}}

= (- 1)^{3} \frac{72 s ( - s ^{2} + 9 )}{( s ^{2} + 9 ) ^{4}}

= - \frac{72 s ( - s ^{2} + 9 )}{( s ^{2} + 9 ) ^{4}}

\frac{\partial}{\partial t} (cos (2 t))

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

x = y^{2} - 1, x = 1 - y^{2}

12 x^{2} - 48 x

\frac{d y}{d x} = y + 3

What is laplacetransform t^3*sin(3t) ?
The answer to laplacetransform t^3*sin(3t) is -(72s(-s^2+9))/((s^2+9)^4)