What is laplacetransform t^2sin(4t) ?

The answer to laplacetransform t^2sin(4t) is -(8(-3s^2+16))/((s^2+16)^3)

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laplacetransform t^2sin(4t)

laplace transform

t^{2} sin (4 t)

- \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}}

Solution steps

L {t^{2} sin (4 t)}

Use Laplace Transform table:

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

For

t^{2} sin (4 t) : f (t) = sin (4 t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {sin (4 t)})

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

\frac{d ^{2}}{d s ^{2}} (\frac{4}{s ^{2} + 16}) = - \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}}

= (- 1)^{2} (- \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}})

= - \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}}

\frac{d y}{d t} + e^{t} y = 12 e^{t}

\frac{x}{7} + \frac{7}{x}

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

\int (sin (θ)) d θ

3 x (18 x^{4} + \frac{13}{x + 1})

What is laplacetransform t^2sin(4t) ?
The answer to laplacetransform t^2sin(4t) is -(8(-3s^2+16))/((s^2+16)^3)