inverselaplace 1/(s^4(s+3))

解答

的拉普拉斯逆变换

\frac{1}{s ^{4} ( s + 3 )}

- \frac{1}{81} H (t) + \frac{t}{27} - \frac{t ^{2}}{18} + \frac{t ^{3}}{18} + \frac{1}{81} e^{- 3 t}

求解步骤

L^{- 1} {\frac{1}{s ^{4} ( s + 3 )}}

将

\frac{1}{s ^{4} ( s + 3 )}

用部份分式展开:

- \frac{1}{81 s} + \frac{1}{27 s ^{2}} - \frac{1}{9 s ^{3}} + \frac{1}{3 s ^{4}} + \frac{1}{81 ( s + 3 )}

= L^{- 1} {- \frac{1}{81 s} + \frac{1}{27 s ^{2}} - \frac{1}{9 s ^{3}} + \frac{1}{3 s ^{4}} + \frac{1}{81 ( s + 3 )}}

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}{81 s}} + L^{- 1} {\frac{1}{27 s ^{2}}} - L^{- 1} {\frac{1}{9 s ^{3}}} + L^{- 1} {\frac{1}{3 s ^{4}}} + L^{- 1} {\frac{1}{81 ( s + 3 )}}

L^{- 1} {\frac{1}{81 s}} : \frac{1}{81} H (t)

L^{- 1} {\frac{1}{27 s ^{2}}} : \frac{t}{27}

L^{- 1} {\frac{1}{9 s ^{3}}} : \frac{t ^{2}}{18}

L^{- 1} {\frac{1}{3 s ^{4}}} : \frac{t ^{3}}{18}

L^{- 1} {\frac{1}{81 ( s + 3 )}} : \frac{1}{81} e^{- 3 t}

= - \frac{1}{81} H (t) + \frac{t}{27} - \frac{t ^{2}}{18} + \frac{t ^{3}}{18} + \frac{1}{81} e^{- 3 t}

\frac{\partial}{\partial z} (\frac{z}{x ^{2} + y ^{2} + z ^{2}})

\frac{d y}{d x} + 15 x^{2} y = 45 x^{2}

\frac{d}{d x} (\frac{1}{1 - 5 x})

\int \frac{5}{x} + 5 x d x

\frac{d y}{d x} = 4 (x^{2} + 1)