zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

inverselaplace (s^3+s+2)/(s^4-4)

解答

的拉普拉斯逆变换

\frac{s ^{3} + s + 2}{s ^{4} - 4}

解答

\frac{1}{4} cos (2 t) - \frac{sin ( 2 t )}{2 2} + \frac{3 - 2}{8} e^{- 2 t} + \frac{3 + 2}{8} e^{2 t}

求解步骤

L^{- 1} {\frac{s ^{3} + s + 2}{s ^{4} - 4}}

将

\frac{s ^{3} + s + 2}{s ^{4} - 4}

用部份分式展开:

\frac{s - 2}{4 ( s ^{2} + 2 )} + \frac{3 - 2}{8 ( s + 2 )} + \frac{3 + 2}{8 ( s - 2 )}

= L^{- 1} {\frac{s - 2}{4 ( s ^{2} + 2 )} + \frac{3 - 2}{8 ( s + 2 )} + \frac{3 + 2}{8 ( s - 2 )}}

乘开

= L^{- 1} {\frac{s}{4 ( s ^{2} + 2 )} + \frac{3}{8 ( s + 2 )} + \frac{3}{8 ( s - 2 )} + \frac{2}{8 ( s - 2 )} - \frac{2}{4 ( s ^{2} + 2 )} - \frac{2}{8 ( s + 2 )}}

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{s}{s ^{2} + 2}} - \frac{2}{4} L^{- 1} {\frac{1}{s ^{2} + 2}} + \frac{3}{8} L^{- 1} {\frac{1}{s + 2}} - \frac{2}{8} L^{- 1} {\frac{1}{s + 2}} + \frac{3}{8} L^{- 1} {\frac{1}{s - 2}} + \frac{2}{8} L^{- 1} {\frac{1}{s - 2}}

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

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

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

L^{- 1} {\frac{1}{s - 2}} : e^{2 t}

= \frac{1}{4} cos (2 t) - \frac{2}{4} \cdot \frac{1}{2} sin (2 t) + \frac{3}{8} e^{- 2 t} - \frac{2}{8} e^{- 2 t} + \frac{3}{8} e^{2 t} + \frac{2}{8} e^{2 t}

整理

\frac{1}{4} cos (2 t) - \frac{2}{4} \frac{1}{2} sin (2 t) + \frac{3}{8} e^{- 2 t} - \frac{2}{8} e^{- 2 t} + \frac{3}{8} e^{2 t} + \frac{2}{8} e^{2 t} : \frac{1}{4} cos (2 t) - \frac{sin ( 2 t )}{2 2} + \frac{3 - 2}{8} e^{- 2 t} + \frac{3 + 2}{8} e^{2 t}

= \frac{1}{4} cos (2 t) - \frac{sin ( 2 t )}{2 2} + \frac{3 - 2}{8} e^{- 2 t} + \frac{3 + 2}{8} e^{2 t}

流行的例子

(2xy^2+3x^2)dx+(2x^2y+4y^3)dy=0

(2 x y^{2} + 3 x^{2}) d x + (2 x^{2} y + 4 y^{3}) d y = 0

(\partial)/(\partial x)(2sqrt(x)+3sqrt(z))

\frac{\partial}{\partial x} (2 x + 3 z)

sum from n=0 to infinity of 1/2 n

n = 0 \sum \infty \frac{1}{2} n

integral of (x^2-x-5)/(x+2)

\int \frac{x ^{2} - x - 5}{x + 2} d x

integral of (x^2+3)e^{-x}

\int (x^{2} + 3) e^{- x} d x