What is laplacetransform te^{-2t}cos(3t) ?

The answer to laplacetransform te^{-2t}cos(3t) is -(-s^2-4s+5)/((s^2+4s+13)^2)

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laplacetransform te^{-2t}cos(3t)

laplace transform

t e^{- 2 t} cos (3 t)

- \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

Solution steps

L {e^{- 2 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^{- 2 t} cos (3 t) : f (t) = e^{- 2 t} cos (3 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 2 t} cos (3 t)})

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

\frac{d}{d s} (\frac{s + 2}{( s + 2 ) ^{2} + 9}) = \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

= - \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

\int \frac{1}{x ^{2} - 0.01} d x

\int \frac{x ^{3}}{1 + x} d x

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 4 t^{2}

\int_{- 2}^{4} ∣ - 2 x + 4 ∣ d x

y^{^{'}} - \frac{2}{x} y = 0

What is laplacetransform te^{-2t}cos(3t) ?
The answer to laplacetransform te^{-2t}cos(3t) is -(-s^2-4s+5)/((s^2+4s+13)^2)