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受欢迎的微积分 >

inverselaplace s/(s^4+4)

解答

的拉普拉斯逆变换

\frac{s}{s ^{4} + 4}

解答

- \frac{1}{4} e^{- t} sin (t) + \frac{1}{4} e^{t} sin (t)

求解步骤

L^{- 1} {\frac{s}{s ^{4} + 4}}

将

\frac{s}{s ^{4} + 4}

用部份分式展开:

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

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

展开

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

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

展开

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

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

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

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

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

= - \frac{1}{4} e^{- t} sin (t) + \frac{1}{4} e^{t} sin (t)

流行的例子

derivative of 20cos(x)

\frac{d}{d x} (20 cos (x))

(\partial)/(\partial x)(5x^2+y^2)

\frac{\partial}{\partial x} (5 x^{2} + y^{2})

integral of 4xln(7x)

\int 4 x ln (7 x) d x

(dv)/(dt)=9.8-0.6

\frac{d v}{d t} = 9.8 - 0.6

integral from 1 to 4 of 2pix(-3/2 x+6)

\int_{1}^{4} 2 π x (- \frac{3}{2} x + 6) d x