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人気のある微分積分 >

laplacetransform sin(4t+5pi)

解

ラプラス変換

sin (4 t + 5 π)

解

- \frac{4}{s ^{2} + 16}

解答ステップ

L {sin (4 t + 5 π)}

ラプラス変換表を使用する:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (4 t + 5 π)} = \frac{s sin ( 5 π ) + 4 cos ( 5 π )}{s ^{2} + 4 ^{2}}

= \frac{s sin ( 5 π ) + 4 cos ( 5 π )}{s ^{2} + 4 ^{2}}

= - \frac{4}{s ^{2} + 16}

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