What is laplacetransform te^{-3t}cos(3t) ?

The answer to laplacetransform te^{-3t}cos(3t) is -(-s^2-6s)/((s^2+6s+18)^2)

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laplacetransform te^{-3t}cos(3t)

laplace transform

t e^{- 3 t} cos (3 t)

- \frac{- s ^{2} - 6 s}{( s ^{2} + 6 s + 18 ) ^{2}}

Solution steps

L {e^{- 3 t} t cos (3 t)}

Use Laplace Transform table:

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

For

t e^{- 3 t} cos (3 t) : f (t) = e^{- 3 t} cos (3 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 3 t} cos (3 t)})

L {e^{- 3 t} cos (3 t)} : \frac{s + 3}{( s + 3 ) ^{2} + 9}

\frac{d}{d s} (\frac{s + 3}{( s + 3 ) ^{2} + 9}) = \frac{- s ^{2} - 6 s}{( s ^{2} + 6 s + 18 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 6 s}{( s ^{2} + 6 s + 18 ) ^{2}}

= - \frac{- s ^{2} - 6 s}{( s ^{2} + 6 s + 18 ) ^{2}}

\frac{d}{d x} (\frac{80 - x}{4})

\int \frac{7 x ^{3} - x ^{2} + 8 x + 12}{x ^{4} + 4 x ^{2}} d x

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

\frac{d}{d x} (sin^{2} (x^{3} + 4 x^{2} - 3))

x \to 0 - lim (\frac{x ^{2} - x}{x - 1})

What is laplacetransform te^{-3t}cos(3t) ?
The answer to laplacetransform te^{-3t}cos(3t) is -(-s^2-6s)/((s^2+6s+18)^2)