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

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

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laplacetransform t^2*sin(1/2 t)

laplace transform

t^{2} \cdot 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}}

\int x^{2} e^{4 x + 3} d x

\int \frac{( x + 6 )}{5 - 4 x - x ^{2}} d x

9 - x

f (x) = (\frac{4 x}{7 x + 5})^{4}

y^{^{''}} + y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = 1

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