inverselaplace 1/(s(s^2+81))

解答

的拉普拉斯逆变换

\frac{1}{s ( s ^{2} + 81 )}

\frac{1}{81} H (t) - \frac{1}{81} cos (9 t)

求解步骤

L^{- 1} {\frac{1}{s ( s ^{2} + 81 )}}

将

\frac{1}{s ( s ^{2} + 81 )}

用部份分式展开:

\frac{1}{81 s} - \frac{s}{81 ( s ^{2} + 81 )}

= L^{- 1} {\frac{1}{81 s} - \frac{s}{81 ( s ^{2} + 81 )}}

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{s}{81 ( s ^{2} + 81 )}}

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

L^{- 1} {\frac{s}{81 ( s ^{2} + 81 )}} : \frac{1}{81} cos (9 t)

= \frac{1}{81} H (t) - \frac{1}{81} cos (9 t)

\frac{d}{d x} (\frac{d y}{d x}) + 81 y = 0

\int cos (\frac{u}{v}) d u

x \to + 0 lim (\frac{sin ( 0 )}{0})

l a pl a ce t r an s f or m 4 e^{(\frac{2 t}{3})}

\int \frac{1}{x ^{3} x ^{2} - 25} d x