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受欢迎的微积分 >

inverselaplace 1/((s^2+4)(s^2+2s+2))

解答

的拉普拉斯逆变换

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

解答

- \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

求解步骤

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

将

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

用部份分式展开:

\frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{s + 3}{10 ( s ^{2} + 2 s + 2 )}

= L^{- 1} {\frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{s + 3}{10 ( s ^{2} + 2 s + 2 )}}

展开

\frac{s + 3}{10 ( s ^{2} + 2 s + 2 )} : \frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1}

= L^{- 1} {\frac{- s - 1}{10 ( s ^{2} + 4 )} + \frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1}}

乘开

= L^{- 1} {- \frac{s}{10 ( s ^{2} + 4 )} + \frac{1}{10} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 1} + \frac{1}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 1} - \frac{1}{10 ( s ^{2} + 4 )}}

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)}

= - \frac{1}{10} L^{- 1} {\frac{s}{s ^{2} + 4}} - \frac{1}{10} L^{- 1} {\frac{1}{s ^{2} + 4}} + \frac{1}{10} L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 1}} + \frac{1}{5} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}}

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 1}} : e^{- t} sin (t)

= - \frac{1}{10} cos (2 t) - \frac{1}{10} \cdot \frac{1}{2} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

整理

- \frac{1}{10} cos (2 t) - \frac{1}{10} \frac{1}{2} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t) : - \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

= - \frac{1}{10} cos (2 t) - \frac{1}{20} sin (2 t) + \frac{1}{10} e^{- t} cos (t) + \frac{1}{5} e^{- t} sin (t)

流行的例子

laplacetransform 2t*sin(t)

l a pl a ce t r an s f or m 2 t \cdot sin (t)

ty^'-y=(-t^3)/(y^2)

t y^{^{'}} - y = \frac{- t ^{3}}{y ^{2}}

7x((dy)/(dx))=2xe^x-7y+6x^2

7 x (\frac{d y}{d x}) = 2 x e^{x} - 7 y + 6 x^{2}

limit as x approaches 2 of (4x-3)(x^2+5)

x \to 2 lim ((4 x - 3) (x^{2} + 5))

derivative arcsin(2/(x^2))

d er i v a t i v e arcsin (\frac{2}{x ^{2}})