What is laplacetransform f(t)=te^{-3t}cos(5t) ?

The answer to laplacetransform f(t)=te^{-3t}cos(5t) is -(-s^2-6s+16)/((s^2+6s+34)^2)

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laplacetransform f(t)=te^{-3t}cos(5t)

laplace transform

f (t) = t e^{- 3 t} cos (5 t)

- \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

Solution steps

L {e^{- 3 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^{- 3 t} cos (5 t) : f (t) = e^{- 3 t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 3 t} cos (5 t)})

L {e^{- 3 t} cos (5 t)} : \frac{s + 3}{( s + 3 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 3}{( s + 3 ) ^{2} + 25}) = \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

= - \frac{- s ^{2} - 6 s + 16}{( s ^{2} + 6 s + 34 ) ^{2}}

\frac{d}{d x} (v) = \frac{1}{2 v}

\int (4 - x) 8^{(4 - x)^{2}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'}} - x e^{y} = 2 e^{y}

\frac{d y}{d x} + 5 y = 3

What is laplacetransform f(t)=te^{-3t}cos(5t) ?
The answer to laplacetransform f(t)=te^{-3t}cos(5t) is -(-s^2-6s+16)/((s^2+6s+34)^2)