inverselaplace 1/((s^2-2s))

解答

的拉普拉斯逆变换

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

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

求解步骤

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

将

\frac{1}{s ^{2} - 2 s}

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_{0}^{5} \frac{1}{x ^{2} - 6 x + 5} d x

x \to 3 + lim (ln x^{2} - 9)

\frac{d ^{3}}{d x ^{3}} (6 x + 8)

l a pl a ce t r an s f or m (1 - e^{t} + 3 e^{- 4 t}) cos (5 t)

\int \frac{x sin ( x ) + cos ( x ) - 1}{x ^{2}} d x