laplacetransform cos^3(t)

解答

拉普拉斯变换

cos^{3} (t)

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

求解步骤

L {cos^{3} (t)}

使用三角恒等式改写

L {\frac{1}{4} cos (3 t) + \frac{3}{4} cos (t)}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

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

= \frac{1}{4} L {cos (3 t)} + \frac{3}{4} L {cos (t)}

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

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

= \frac{1}{4} \cdot \frac{s}{s ^{2} + 9} + \frac{3}{4} \cdot \frac{s}{s ^{2} + 1}

整理

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

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

\frac{d y}{d x} = \frac{6 y e ^{3 x}}{2 + e ^{3 x}}

\int_{- \infty}^{- 1} e^{- 28 t} d t

\frac{d}{d x} (x (x - 1)^{2} (x - 2)^{3})

\int π^{π} d x

d er i v a t i v e y = x (3 x + 4)^{5}