laplacetransform sin(2t)cos(3t)

解

ラプラス変換

sin (2 t) cos (3 t)

\frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}

解答ステップ

L {sin (2 t) cos (3 t)}

三角関数の公式を使用して書き換える

L {\frac{1}{2} sin (5 t) - \frac{1}{2} sin (t)}

ラプラス変形の線形性を使用する:

関数

f (t), g (t)

と定数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= \frac{1}{2} L {sin (5 t)} - \frac{1}{2} L {sin (t)}

L {sin (5 t)} : \frac{5}{s ^{2} + 25}

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

= \frac{1}{2} \cdot \frac{5}{s ^{2} + 25} - \frac{1}{2} \cdot \frac{1}{s ^{2} + 1}

改良

\frac{1}{2} \frac{5}{s ^{2} + 25} - \frac{1}{2} \frac{1}{s ^{2} + 1} : \frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}

= \frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}