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

解答

拉普拉斯变换

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

- \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

求解步骤

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

使用拉普拉斯变换表:

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

对于

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

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

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

\frac{d ^{2}}{d s ^{2}} (\frac{2}{( s + 2 ) ^{2} + 4}) = - \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

= (- 1)^{2} (- \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}})

= - \frac{4 ( - 3 s ^{2} - 12 s - 8 )}{( s ^{2} + 4 s + 8 ) ^{3}}

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