laplacetransform sin^2(7t)

解答

拉普拉斯变换

sin^{2} (7 t)

\frac{98}{s ( s ^{2} + 196 )}

求解步骤

L {sin^{2} (7 t)}

使用三角恒等式改写

L {\frac{1}{2} - \frac{1}{2} cos (14 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)}

= L {\frac{1}{2}} - \frac{1}{2} L {cos (14 t)}

L {\frac{1}{2}} : \frac{1}{2 s}

L {cos (14 t)} : \frac{s}{s ^{2} + 196}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 196}

化简

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 196} : \frac{98}{s ( s ^{2} + 196 )}

= \frac{98}{s ( s ^{2} + 196 )}

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

\int cos^{16} (x) sin (x) d x

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

\frac{\partial}{\partial x} (x^{2} - x - 12 y + 1)

\frac{d}{d x} (524.8 \cdot x^{0.75})