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

The answer to laplacetransform t^2cos(4t) is -(2s(-s^2+48))/((s^2+16)^3)

laplacetransform t^2cos(4t)

laplace transform

t^{2} cos (4 t)

- \frac{2 s ( - s ^{2} + 48 )}{( s ^{2} + 16 ) ^{3}}

Solution steps

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

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{s}{s ^{2} + 16}) = - \frac{2 s ( - s ^{2} + 48 )}{( s ^{2} + 16 ) ^{3}}

= (- 1)^{2} (- \frac{2 s ( - s ^{2} + 48 )}{( s ^{2} + 16 ) ^{3}})

= - \frac{2 s ( - s ^{2} + 48 )}{( s ^{2} + 16 ) ^{3}}

\int ye d y

\int cos (x) 9 - sin^{2} (x) d x

7 x y^{^{'}} - 2 y = \frac{- x ^{2}}{y ^{6}}

\frac{d}{d x} (5 x^{2} + 3 x)

y^{^{''}} - 2 y^{^{'}} + y = 12 t e^{t} + 2 e^{t} + t - 4

What is laplacetransform t^2cos(4t) ?
The answer to laplacetransform t^2cos(4t) is -(2s(-s^2+48))/((s^2+16)^3)