inverselaplace s/((s^2-1^2)(s^2+17))

解答

的拉普拉斯逆变换

\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )}

\frac{1}{36} e^{- t} + \frac{1}{36} e^{t} - \frac{1}{18} cos (17 t)

求解步骤

L^{- 1} {\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )}}

将

\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )}

用部份分式展开:

\frac{1}{36 ( s + 1 )} + \frac{1}{36 ( s - 1 )} - \frac{s}{18 ( s ^{2} + 17 )}

= L^{- 1} {\frac{1}{36 ( s + 1 )} + \frac{1}{36 ( s - 1 )} - \frac{s}{18 ( s ^{2} + 17 )}}

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}{36 ( s + 1 )}} + L^{- 1} {\frac{1}{36 ( s - 1 )}} - L^{- 1} {\frac{s}{18 ( s ^{2} + 17 )}}

L^{- 1} {\frac{1}{36 ( s + 1 )}} : \frac{1}{36} e^{- t}

L^{- 1} {\frac{1}{36 ( s - 1 )}} : \frac{1}{36} e^{t}

L^{- 1} {\frac{s}{18 ( s ^{2} + 17 )}} : \frac{1}{18} cos (17 t)

= \frac{1}{36} e^{- t} + \frac{1}{36} e^{t} - \frac{1}{18} cos (17 t)

x \to - \infty lim (e^{x} \cdot x)

\int \frac{x + 10}{x ^{2} + 6 x + 13} d x

\frac{\partial}{\partial x} (- x + y)

\int x^{2} tan^{3} (x^{3}) d x

(x^{2} + 4) y^{^{'}} + 3 x y = x, y (0) = 1