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

The answer to laplacetransform t^2sin(3t) is -(18(-s^2+3))/((s^2+9)^3)

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laplacetransform t^2sin(3t)

laplace transform

t^{2} sin (3 t)

- \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}}

Solution steps

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

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{3}{s ^{2} + 9}) = - \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}}

= (- 1)^{2} (- \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}})

= - \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}}

\int_{- 3}^{3} 12 - x^{2} - (x^{2} - 6) d x

x \to 2 lim (9 + (- 12))

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2}})

x \to 0 lim (\frac{x ^{2}}{sin ^{2} ( 7 x )})

f (x) = 4 sec (x)

What is laplacetransform t^2sin(3t) ?
The answer to laplacetransform t^2sin(3t) is -(18(-s^2+3))/((s^2+9)^3)