What is laplacetransform t^2e^{-3t}sin(4t) ?

The answer to laplacetransform t^2e^{-3t}sin(4t) is -(8(-3s^2-18s-11))/((s^2+6s+25)^3)

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laplacetransform t^2e^{-3t}sin(4t)

laplace transform

t^{2} e^{- 3 t} sin (4 t)

- \frac{8 ( - 3 s ^{2} - 18 s - 11 )}{( s ^{2} + 6 s + 25 ) ^{3}}

Solution steps

L {e^{- 3 t} t^{2} sin (4 t)}

Use Laplace Transform table:

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

For

t^{2} e^{- 3 t} sin (4 t) : f (t) = e^{- 3 t} sin (4 t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {e^{- 3 t} sin (4 t)})

L {e^{- 3 t} sin (4 t)} : \frac{4}{( s + 3 ) ^{2} + 16}

\frac{d ^{2}}{d s ^{2}} (\frac{4}{( s + 3 ) ^{2} + 16}) = - \frac{8 ( - 3 s ^{2} - 18 s - 11 )}{( s ^{2} + 6 s + 25 ) ^{3}}

= (- 1)^{2} (- \frac{8 ( - 3 s ^{2} - 18 s - 11 )}{( s ^{2} + 6 s + 25 ) ^{3}})

= - \frac{8 ( - 3 s ^{2} - 18 s - 11 )}{( s ^{2} + 6 s + 25 ) ^{3}}

\frac{d}{d x} ((\frac{x ^{2} + 4}{x ^{2} - 4})^{3})

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

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

x \to - 4 lim (x^{2} + 3 x - 7)

What is laplacetransform t^2e^{-3t}sin(4t) ?
The answer to laplacetransform t^2e^{-3t}sin(4t) is -(8(-3s^2-18s-11))/((s^2+6s+25)^3)