What is laplacetransform tcos(2t) ?

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

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laplacetransform tcos(2t)

laplace transform

t cos (2 t)

- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

Solution steps

L {t cos (2 t)}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t cos (2 t) : f (t) = cos (2 t), k = 1

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

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

\frac{d}{d s} (\frac{s}{s ^{2} + 4}) = \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

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

= - \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

k = 0 \sum \infty (\frac{e}{10})^{k}

x \to 1 - lim (- x^{2} + 3 x - 4)

f (t) = 4 e^{- 7 t} cos (9 t)

\frac{d}{d x} (e^{x} (cos (x)))

\frac{\partial}{\partial y} (4 x^{2} y^{3} + 4)

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