laplacetransform e^{3t}(9-4t+10sin(t/2))

解

ラプラス変換

e^{3 t} (9 - 4 t + 10 sin (\frac{t}{2}))

\frac{9}{s - 3} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}

解答ステップ

L {e^{3 t} (- 4 t + 10 sin (\frac{t}{2}) + 9)}

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^{3 t} (9 - 4 t + 10 sin (\frac{t}{2}))

向け

: f (t) = (9 - 4 t + 10 sin (\frac{t}{2})), a = 3

= L {(9 - 4 t + 10 sin (\frac{t}{2}))} (s - 3)

L {(9 - 4 t + 10 sin (\frac{t}{2}))} : \frac{9}{s} - \frac{4}{s ^{2}} + \frac{20}{4 s ^{2} + 1}

= \frac{9}{( s - 3 )} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}

= \frac{9}{s - 3} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}