What is laplacetransform t^2sin(1/2 t) ?

The answer to laplacetransform t^2sin(1/2 t) is -(16(1-12s^2))/((4s^2+1)^3)

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laplacetransform t^2sin(1/2 t)

laplace transform

t^{2} sin (\frac{1}{2} t)

- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

Solution steps

L {t^{2} sin (\frac{1}{2} 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 (\frac{1}{2} t) : f (t) = sin (\frac{1}{2} t), k = 2

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

L {sin (\frac{1}{2} t)} : \frac{2}{4 s ^{2} + 1}

\frac{d ^{2}}{d s ^{2}} (\frac{2}{4 s ^{2} + 1}) = - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

= (- 1)^{2} (- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}})

= - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

n = 0 \sum \infty n^{n} x^{n}

x \to 0 lim (\frac{3 - 3 cos ( x )}{sin ^{2} ( x )})

y = (x^{2} - 9 x + 8) 25 - x^{2}

f (x) = \frac{11}{( 1 - x ) ^{2}}

\int sin^{3} (2 x) d x

What is laplacetransform t^2sin(1/2 t) ?
The answer to laplacetransform t^2sin(1/2 t) is -(16(1-12s^2))/((4s^2+1)^3)