laplacetransform t^3cos(3t)

解答

拉普拉斯变换

t^{3} cos (3 t)

- \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

求解步骤

L {t^{3} cos (3 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^{3} cos (3 t) : f (t) = cos (3 t), k = 3

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

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

\frac{d ^{3}}{d s ^{3}} (\frac{s}{s ^{2} + 9}) = \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

= (- 1)^{3} \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

= - \frac{6 ( - s ^{4} + 54 s ^{2} - 81 )}{( s ^{2} + 9 ) ^{4}}

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