laplacetransform 1-cos(7t)

解答

拉普拉斯变换

1 - cos (7 t)

\frac{49}{s ( s ^{2} + 49 )}

求解步骤

L {1 - cos (7 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 {1} - L {cos (7 t)}

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

L {cos (7 t)} : \frac{s}{s ^{2} + 49}

= \frac{1}{s} - \frac{s}{s ^{2} + 49}

化简

\frac{1}{s} - \frac{s}{s ^{2} + 49} : \frac{49}{s ( s ^{2} + 49 )}

= \frac{49}{s ( s ^{2} + 49 )}

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

x \to 0 lim ((e^{x} + 1)^{\frac{1}{x}})

a re a x = y (y - 13), x = - y (y - 13)

\frac{d}{d x} (e^{x} sin (x) + cos (x) e^{x})

\frac{d}{d x} (\frac{- 6 x}{( x ^{2} - 4 ) ^{2}})