inverselaplace 1/(s^2+9s+9)

解

逆ラプラス

\frac{1}{s ^{2} + 9 s + 9}

e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})

解答ステップ

L^{- 1} {\frac{1}{s ^{2} + 9 s + 9}}

\frac{1}{s ^{2} + 9 s + 9} = \frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}

= L^{- 1} {\frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}}

逆変換ルールを適用:

L^{- 1} {F (s)} = f (t)

ならば

L^{- 1} {F (s - a)} = e^{a t} f (t)

\frac{1}{( s + \frac{9}{2} ) ^{2} - \frac{45}{4}}

向け

: a = - \frac{9}{2}, F (s) = \frac{1}{s ^{2} - \frac{45}{4}}

= e^{- \frac{9}{2} t} L^{- 1} {\frac{1}{s ^{2} - \frac{45}{4}}}

L^{- 1} {\frac{1}{s ^{2} - \frac{45}{4}}} : \frac{2}{3 5} sinh (\frac{3 5 t}{2})

= e^{- \frac{9 t}{2}} \frac{2}{3 5} sinh (\frac{3 5 t}{2})