inverselaplace 1/(s(s+2))

解答

的拉普拉斯逆变换

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

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

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

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

\int 13 (tan (x)) (ln (cos (x))) d x

x \to + \frac{π}{2} lim (cot (x))

d er i v a t i v e f (x) = 5 arcsin (x^{3})

d er i v a t i v e \frac{e ^{x}}{x}

\int_{1}^{2} x ln (x) d x