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受欢迎的微积分 >

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

解答

的拉普拉斯逆变换

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

解答

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

求解步骤

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

将

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

用部份分式展开:

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

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

乘开

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

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

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

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

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

整理

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

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

流行的例子

integral of (sqrt(x)+\sqrt[3]{x})

\int (x + 3 x) d x

limit as x approaches 2 of x^2-7x+2

x \to 2 lim (x^{2} - 7 x + 2)

tangent f(x)=3x^3+8x^2,\at x=-2

t an g e n t f (x) = 3 x^{3} + 8 x^{2}, at x = - 2

limit as x approaches 5 of 7/(x-5)

x \to 5 lim (\frac{7}{x - 5})

limit as x approaches infinity of 7+4/x

x \to \infty lim (7 + \frac{4}{x})