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

The answer to laplacetransform te^{-3t}cos(2t) is -(-s^2-6s-5)/((s^2+6s+13)^2)

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

laplace transform

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

- \frac{- s ^{2} - 6 s - 5}{( s ^{2} + 6 s + 13 ) ^{2}}

Solution steps

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

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

L {e^{- 3 t} cos (2 t)} : \frac{s + 3}{( s + 3 ) ^{2} + 4}

\frac{d}{d s} (\frac{s + 3}{( s + 3 ) ^{2} + 4}) = \frac{- s ^{2} - 6 s - 5}{( s ^{2} + 6 s + 13 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 6 s - 5}{( s ^{2} + 6 s + 13 ) ^{2}}

= - \frac{- s ^{2} - 6 s - 5}{( s ^{2} + 6 s + 13 ) ^{2}}

\int_{1}^{\infty} \frac{1}{4 ^{x}} d x

\int 3 - 3 x d x

\int x cos (2 x) d x

y = e^{x^{6}}

\frac{d y}{d x} + \frac{1}{3} y = e^{x} y^{4}

What is laplacetransform te^{-3t}cos(2t) ?
The answer to laplacetransform te^{-3t}cos(2t) is -(-s^2-6s-5)/((s^2+6s+13)^2)