inverselaplace (36)/(s(s^2+1)(s^2+9))

解答

的拉普拉斯逆变换

\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )}

4 H (t) - \frac{9}{2} cos (t) + \frac{1}{2} cos (3 t)

求解步骤

L^{- 1} {\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )}}

将

\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )}

用部份分式展开:

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

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

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

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

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

L^{- 1} {\frac{s}{2 ( s ^{2} + 9 )}} : \frac{1}{2} cos (3 t)

= 4 H (t) - \frac{9}{2} cos (t) + \frac{1}{2} cos (3 t)

\int e^{- 23 x} d x

t y^{^{'}} = 3 y + t^{4} sin (4 t)

d er i v a t i v e f (x) = \frac{1}{x + 2}

\int \frac{4 x}{x ^{2} - 9} d x

\int 4 e^{2 x} sin (2 x) d x