laplacetransform t^2sin(4t)

解

ラプラス変換

t^{2} sin (4 t)

- \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}}

解答ステップ

L {t^{2} sin (4 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^{2} sin (4 t)

向け

: f (t) = sin (4 t), k = 2

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

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

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

= (- 1)^{2} (- \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}})

= - \frac{8 ( - 3 s ^{2} + 16 )}{( s ^{2} + 16 ) ^{3}}