laplacetransform sin(1/2 t)*t^2

解

ラプラス変換

sin (\frac{1}{2} t) \cdot t^{2}

- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

解答ステップ

L {t^{2} sin (\frac{1}{2} 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)})

sin (\frac{1}{2} t) t^{2}

向け

: f (t) = sin (\frac{1}{2} t), k = 2

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

L {sin (\frac{1}{2} t)} : \frac{2}{4 s ^{2} + 1}

\frac{d ^{2}}{d s ^{2}} (\frac{2}{4 s ^{2} + 1}) = - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

= (- 1)^{2} (- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}})

= - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}