inverselaplace {(4s)/(4s^2+1)}

解答

的拉普拉斯逆变换

{\frac{4 s}{4 s ^{2} + 1}}

cos (\frac{t}{2})

求解步骤

L^{- 1} {\frac{4 s}{4 s ^{2} + 1}}

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

对于

\frac{4 s}{4 s ^{2} + 1} : F (s) = \frac{4}{4 s ^{2} + 1}

= \frac{d}{d t} (L^{- 1} {\frac{4}{4 s ^{2} + 1}}) + L^{- 1} {\frac{4}{4 s ^{2} + 1}} (0)

L^{- 1} {\frac{4}{4 s ^{2} + 1}} : 2 sin (\frac{t}{2})

\frac{d}{d t} (2 sin (\frac{t}{2})) = cos (\frac{t}{2})

在点

0 = 0

处计算

2 sin (\frac{t}{2})

= cos (\frac{t}{2})

x \to 0 lim (2 x - x^{2} - x^{3})

\int cos (x^{5}) d x

in v erse l a pl a ce \frac{3}{s ^{2} - 25}

d er i v a t i v e sin (x) + \frac{7}{4} cot (x)

s l o p e \frac{x ^{2}}{x + 7}