inverselaplace s/((s^2+4))

解答

的拉普拉斯逆变换

\frac{s}{( s ^{2} + 4 )}

cos (2 t)

求解步骤

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

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

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

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

\frac{d}{d t} (\frac{1}{2} sin (2 t)) = cos (2 t)

在点

0 = 0

处计算

\frac{1}{2} sin (2 t)

= cos (2 t)

x \to 1 lim ((x^{2} - 2)^{3})

d er i v a t i v e f (x) = 6 sec (x) tan (x)

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

\int \frac{1}{4 - y ^{2}} d y

y^{^{'}} = sin (y)