laplacetransform t^3cos(t)

解

ラプラス変換

t^{3} cos (t)

- \frac{6 ( - s ^{4} + 6 s ^{2} - 1 )}{( s ^{2} + 1 ) ^{4}}

解答ステップ

L {t^{3} cos (t)}

ラプラス変換表を使用する:

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

t^{3} cos (t)

向け

: f (t) = cos (t), k = 3

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

L {cos (t)} : \frac{s}{s ^{2} + 1}

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

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

= - \frac{6 ( - s ^{4} + 6 s ^{2} - 1 )}{( s ^{2} + 1 ) ^{4}}