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受欢迎的微积分 >

inverselaplace 4/((s+1)(s^2+4s+8))

解答

的拉普拉斯逆变换

\frac{4}{( s + 1 ) ( s ^{2} + 4 s + 8 )}

解答

\frac{4}{5} e^{- t} - \frac{4}{5} e^{- 2 t} cos (2 t) - \frac{2 e ^{- 2 t} sin ( 2 t )}{5}

求解步骤

L^{- 1} {\frac{4}{( s + 1 ) ( s ^{2} + 4 s + 8 )}}

将

\frac{4}{( s + 1 ) ( s ^{2} + 4 s + 8 )}

用部份分式展开:

\frac{4}{5 ( s + 1 )} + \frac{- 4 s - 12}{5 ( s ^{2} + 4 s + 8 )}

= L^{- 1} {\frac{4}{5 ( s + 1 )} + \frac{- 4 s - 12}{5 ( s ^{2} + 4 s + 8 )}}

展开

\frac{- 4 s - 12}{5 ( s ^{2} + 4 s + 8 )} : - \frac{4}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 4} - \frac{4}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 4}

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

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

L^{- 1} {\frac{4}{5 ( s + 1 )}} : \frac{4}{5} e^{- t}

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

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

= \frac{4}{5} e^{- t} - \frac{4}{5} e^{- 2 t} cos (2 t) - \frac{4}{5} e^{- 2 t} \frac{1}{2} sin (2 t)

整理

\frac{4}{5} e^{- t} - \frac{4}{5} e^{- 2 t} cos (2 t) - \frac{4}{5} e^{- 2 t} \frac{1}{2} sin (2 t) : \frac{4}{5} e^{- t} - \frac{4}{5} e^{- 2 t} cos (2 t) - \frac{2 e ^{- 2 t} sin ( 2 t )}{5}

= \frac{4}{5} e^{- t} - \frac{4}{5} e^{- 2 t} cos (2 t) - \frac{2 e ^{- 2 t} sin ( 2 t )}{5}

流行的例子

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

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

(\partial)/(\partial y)(-xsin(y))

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

y^'=(t+y)^2-1,y(1)=6

y^{^{'}} = (t + y)^{2} - 1, y (1) = 6

derivative of 4x+2x^2sin(1/x)

\frac{d}{d x} (4 x + 2 x^{2} sin (\frac{1}{x}))

integral of sin(x-nx)

\int sin (x - n x) d x