What is laplacetransform t^2sin(t)*e^{-2t} ?

The answer to laplacetransform t^2sin(t)*e^{-2t} is -(2(-3s^2-12s-11))/((s^2+4s+5)^3)

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laplacetransform t^2sin(t)*e^{-2t}

laplace transform

t^{2} sin (t) \cdot e^{- 2 t}

- \frac{2 ( - 3 s ^{2} - 12 s - 11 )}{( s ^{2} + 4 s + 5 ) ^{3}}

Solution steps

L {e^{- 2 t} t^{2} sin (t)}

Use Laplace Transform table:

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

For

t^{2} sin (t) e^{- 2 t} : f (t) = sin (t) e^{- 2 t}, k = 2

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{1}{( s + 2 ) ^{2} + 1}) = - \frac{2 ( - 3 s ^{2} - 12 s - 11 )}{( s ^{2} + 4 s + 5 ) ^{3}}

= (- 1)^{2} (- \frac{2 ( - 3 s ^{2} - 12 s - 11 )}{( s ^{2} + 4 s + 5 ) ^{3}})

= - \frac{2 ( - 3 s ^{2} - 12 s - 11 )}{( s ^{2} + 4 s + 5 ) ^{3}}

\frac{d}{d x} ((1 + x y) e^{x y})

(3 x + 28)^{\frac{1}{2}}

\int e^{\frac{y}{x}} d y

f (x) = 9 - ln (x), at x = 1

\int \frac{1}{x 25 x ^{2} + 1} d x

What is laplacetransform t^2sin(t)*e^{-2t} ?
The answer to laplacetransform t^2sin(t)*e^{-2t} is -(2(-3s^2-12s-11))/((s^2+4s+5)^3)