What is laplacetransform t*e^{-4t}cos(5t) ?

The answer to laplacetransform t*e^{-4t}cos(5t) is -(-s^2-8s+9)/((s^2+8s+41)^2)

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laplacetransform t*e^{-4t}cos(5t)

laplace transform

t \cdot e^{- 4 t} cos (5 t)

- \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

Solution steps

L {e^{- 4 t} t cos (5 t)}

Use Laplace Transform table:

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

For

t e^{- 4 t} cos (5 t) : f (t) = e^{- 4 t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 4 t} cos (5 t)})

L {e^{- 4 t} cos (5 t)} : \frac{s + 4}{( s + 4 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 4}{( s + 4 ) ^{2} + 25}) = \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

= - \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

cot^{2} (x)

x^{\frac{2}{3}} (1 - x)

y = \frac{1}{x}, (1, 4)

\frac{d}{d x} (- e^{x} sin (y))

\int ln^{2} (x + x^{2} + 1) d x

What is laplacetransform t*e^{-4t}cos(5t) ?
The answer to laplacetransform t*e^{-4t}cos(5t) is -(-s^2-8s+9)/((s^2+8s+41)^2)