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

解答

的拉普拉斯逆变换

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

- \frac{1}{4} e^{- t} - \frac{e ^{- t} t}{2} + \frac{1}{4} e^{t}

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

= - \frac{1}{4} e^{- t} - \frac{e ^{- t} t}{2} + \frac{1}{4} e^{t}

h \to 0 lim (\frac{8 ^{h} - 1}{h})

d er i v a t i v e f (x) = \frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}}

\int_{- \infty}^{0} \frac{1}{6 - 4 x} d x

y^{^{''}} + 7 y^{^{'}} + 14 y = 1096 e^{2 t} cos (8 t)

x \to 4 - lim (\frac{x}{x ^{2} - 16})