laplacetransform e^{5t}(-2cos(t)+12sin(t))

解

ラプラス変換

e^{5 t} (- 2 cos (t) + 12 sin (t))

- \frac{2 ( s - 5 )}{( s - 5 ) ^{2} + 1} + \frac{12}{( s - 5 ) ^{2} + 1}

解答ステップ

L {e^{5 t} (12 sin (t) - 2 cos (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^{5 t} (- 2 cos (t) + 12 sin (t))

向け

: f (t) = (- 2 cos (t) + 12 sin (t)), a = 5

= L {(- 2 cos (t) + 12 sin (t))} (s - 5)

L {(- 2 cos (t) + 12 sin (t))} : - \frac{2 s}{s ^{2} + 1} + \frac{12}{s ^{2} + 1}

= - \frac{2 ( s - 5 )}{( s - 5 ) ^{2} + 1} + \frac{12}{( s - 5 ) ^{2} + 1}