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

解答

的拉普拉斯逆变换

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

- H (t) + t + e^{- t}

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

= - H (t) + t + e^{- t}

\int_{0}^{π} u sin^{2} (u) d u

(1 + x^{4}) d y + (1 + 4 y^{2}) d x = 0

y^{^{'}} - 5 y^{(2)} e^{(- 3 x)} = 0, y (0) = 9

\int \frac{x ^{3}}{x ^{2} + 2 x + 1} d x

t an g e n t y = \frac{( x ^{2} )}{8 + x}