What is laplacetransform tcos(4t) ?

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

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laplacetransform tcos(4t)

laplace transform

t cos (4 t)

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

Solution steps

L {t 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 cos (4 t) : f (t) = cos (4 t), k = 1

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

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

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

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

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

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