inverselaplace f(s)= 2/((4s^2-8s))

解答

的拉普拉斯逆变换

f (s) = \frac{2}{( 4 s ^{2} - 8 s )}

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

求解步骤

L^{- 1} {\frac{2}{( 4 s ^{2} - 8 s )}}

将

\frac{2}{4 s ^{2} - 8 s}

用部份分式展开:

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

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

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

L^{- 1} {\frac{1}{4 ( s - 2 )}} : \frac{1}{4} e^{2 t}

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

\frac{\partial}{\partial x} (x^{0.6} \cdot y^{0.4})

\int 4 x (x^{2} + 30000)^{- \frac{2}{3}} d x

\int - \frac{2}{x} d x

t an g e n t f (x) = 2 x, (2, 2)

in v erse l a pl a ce \frac{2}{s ^{2} ( s ^{2} - 2 s + 5 )}