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

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

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laplacetransform te^{-4t}cos(5t)

laplace transform

t 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}}

- x^{4}, x^{2} - 2 x

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

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

\int 3 e^{8 x} d x

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

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