inverselaplace (s+2)/(2s*(s+1))

解答

的拉普拉斯逆变换

\frac{s + 2}{2 s \cdot ( s + 1 )}

H (t) - \frac{1}{2} e^{- t}

求解步骤

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

将

\frac{s + 2}{2 s ( s + 1 )}

用部份分式展开:

\frac{1}{s} - \frac{1}{2 ( s + 1 )}

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

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - \frac{1}{2} e^{- t}

\frac{d ^{2}}{d x ^{2}} (x y e^{(- x + 3 y)})

\frac{\partial}{\partial z} (3 x y z^{2})

d er i v a t i v e e^{2 x} ln (2 x)

x \to - 2 lim (\frac{8}{x})

\frac{\partial}{\partial x} (ln (4 x - 7 y))