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

The answer to laplacetransform te^{-t}cos(5t) is -(-s^2-2s+24)/((s^2+2s+26)^2)

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

laplace transform

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

- \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

Solution steps

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

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

L {e^{- t} cos (5 t)} : \frac{s + 1}{( s + 1 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 1}{( s + 1 ) ^{2} + 25}) = \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

= - \frac{- s ^{2} - 2 s + 24}{( s ^{2} + 2 s + 26 ) ^{2}}

f (x) = sin (9 x) + cos (2 x)

\frac{d}{d x} (sec (x) - tan (x))

x \to 0 lim (sech^{2} (x))

\frac{\partial}{\partial r} (r sin (t))

\int 5 cos (3 x) d x

What is laplacetransform te^{-t}cos(5t) ?
The answer to laplacetransform te^{-t}cos(5t) is -(-s^2-2s+24)/((s^2+2s+26)^2)