laplacetransform f(t)=t(sin(9t)+e^{-2t})

解

ラプラス変換

f (t) = t (sin (9 t) + e^{- 2 t})

- \frac{- s ^{4} - 18 s ^{3} - 234 s ^{2} - 72 s - 6561}{( s + 2 ) ^{2} ( s ^{2} + 81 ) ^{2}}

解答ステップ

L {t (e^{- 2 t} + sin (9 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 (sin (9 t) + e^{- 2 t})

向け

: f (t) = (sin (9 t) + e^{- 2 t}), k = 1

= (- 1)^{1} \frac{d}{d s} (L {(sin (9 t) + e^{- 2 t})})

L {(sin (9 t) + e^{- 2 t})} : \frac{9}{s ^{2} + 81} + \frac{1}{s + 2}

\frac{d}{d s} (\frac{9}{s ^{2} + 81} + \frac{1}{s + 2}) = \frac{- s ^{4} - 18 s ^{3} - 234 s ^{2} - 72 s - 6561}{( s + 2 ) ^{2} ( s ^{2} + 81 ) ^{2}}

= (- 1)^{1} \frac{- s ^{4} - 18 s ^{3} - 234 s ^{2} - 72 s - 6561}{( s + 2 ) ^{2} ( s ^{2} + 81 ) ^{2}}

= - \frac{- s ^{4} - 18 s ^{3} - 234 s ^{2} - 72 s - 6561}{( s + 2 ) ^{2} ( s ^{2} + 81 ) ^{2}}