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受欢迎的微积分 >

inverselaplace 1/((s-1)(s^2+9))

解答

的拉普拉斯逆变换

\frac{1}{( s - 1 ) ( s ^{2} + 9 )}

解答

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

求解步骤

L^{- 1} {\frac{1}{( s - 1 ) ( s ^{2} + 9 )}}

将

\frac{1}{( s - 1 ) ( s ^{2} + 9 )}

用部份分式展开:

\frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}

= L^{- 1} {\frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}}

乘开

= L^{- 1} {- \frac{s}{10 ( s ^{2} + 9 )} + \frac{1}{10 ( s - 1 )} - \frac{1}{10 ( s ^{2} + 9 )}}

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}{10} L^{- 1} {\frac{1}{s - 1}} - \frac{1}{10} L^{- 1} {\frac{s}{s ^{2} + 9}} - \frac{1}{10} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \cdot \frac{1}{3} sin (3 t)

整理

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \frac{1}{3} sin (3 t) : \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

= \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

流行的例子

面积 2x^2-8x-55,-x^2-2x+50,-5,7

a re a 2 x^{2} - 8 x - 55, - x^{2} - 2 x + 50, - 5, 7

(\partial)/(\partial y)(e^{xz}y^3-ln(x))

\frac{\partial}{\partial y} (e^{x z} y^{3} - ln (x))

integral of x(x+3)^7

\int x (x + 3)^{7} d x

x^2(dy)/(dx)=sqrt(y)(4x+3)

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

(\partial)/(\partial x)(-sin(xy))

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