laplacetransform cos(3t+5)e^{2t}

解

ラプラス変換

cos (3 t + 5) e^{2 t}

\frac{cos ( 5 ) ( s - 2 ) - 3 sin ( 5 )}{( s - 2 ) ^{2} + 9}

解答ステップ

L {e^{2 t} cos (3 t + 5)}

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)

cos (3 t + 5) e^{2 t}

向け

: f (t) = cos (3 t + 5), a = 2

= L {cos (3 t + 5)} (s - 2)

L {cos (3 t + 5)} : \frac{cos ( 5 ) s - 3 sin ( 5 )}{s ^{2} + 9}

= \frac{cos ( 5 ) ( s - 2 ) - 3 sin ( 5 )}{( s - 2 ) ^{2} + 9}