inverselaplace 1/((s-1)(s+4))

解答

的拉普拉斯逆变换

\frac{1}{( s - 1 ) ( s + 4 )}

\frac{1}{5} e^{t} - \frac{1}{5} e^{- 4 t}

求解步骤

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

将

\frac{1}{( s - 1 ) ( s + 4 )}

用部份分式展开:

\frac{1}{5 ( s - 1 )} - \frac{1}{5 ( s + 4 )}

= L^{- 1} {\frac{1}{5 ( s - 1 )} - \frac{1}{5 ( s + 4 )}}

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}{5 ( s - 1 )}} - L^{- 1} {\frac{1}{5 ( s + 4 )}}

L^{- 1} {\frac{1}{5 ( s - 1 )}} : \frac{1}{5} e^{t}

L^{- 1} {\frac{1}{5 ( s + 4 )}} : \frac{1}{5} e^{- 4 t}

= \frac{1}{5} e^{t} - \frac{1}{5} e^{- 4 t}

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