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 (10)/(s^3+2s^2+20s)

解答

的拉普拉斯逆变换

\frac{10}{s ^{3} + 2 s ^{2} + 20 s}

解答

\frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

求解步骤

L^{- 1} {\frac{10}{s ^{3} + 2 s ^{2} + 20 s}}

将

\frac{10}{s ^{3} + 2 s ^{2} + 20 s}

用部份分式展开:

\frac{1}{2 s} + \frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )}

= L^{- 1} {\frac{1}{2 s} + \frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )}}

展开

\frac{- s - 2}{2 ( s ^{2} + 2 s + 20 )} : - \frac{1}{2} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 19} - \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + 19}

= L^{- 1} {\frac{1}{2 s} - \frac{1}{2} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 19} - \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + 19}}

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{1}{2 s}} - \frac{1}{2} L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 19}} - \frac{1}{2} L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 19}}

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

L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 19}} : e^{- t} cos (19 t)

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 19}} : e^{- t} \frac{1}{19} sin (19 t)

= \frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2} e^{- t} \frac{1}{19} sin (19 t)

整理

\frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2} e^{- t} \frac{1}{19} sin (19 t) : \frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

= \frac{1}{2} H (t) - \frac{1}{2} e^{- t} cos (19 t) - \frac{1}{2 19} e^{- t} sin (19 t)

流行的例子

integral from 0 to 1 of 0

\int_{0}^{1} 0 d x

integral from 0 to 1 of 1/(4y-1)

\int_{0}^{1} \frac{1}{4 y - 1} d y

derivative f(x)= 6/x

d er i v a t i v e f (x) = \frac{6}{x}

limit as x approaches 0 of ((2x^2+16))/x

x \to 0 lim (\frac{( 2 x ^{2} + 16 )}{x})

泰勒 x^2e^{x^2}

t a y l or x^{2} e^{x^{2}}