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人気のある微分積分 >

laplacetransform e^{-t}cos(2t)

解

ラプラス変換

e^{- t} cos (2 t)

解

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

解答ステップ

L {e^{- t} cos (2 t)}

変換ルールを適用:

L {f (t)} = F (s)

ならば

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

e^{- t} cos (2 t)

向け

: f (t) = cos (2 t), a = - 1

= L {cos (2 t)} (s + 1)

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

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

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

人気の例

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\frac{\partial}{\partial x} (x^{\frac{1}{2}} y^{\frac{1}{2}} z)

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\int_{0}^{6} 1 + x d x

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