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受欢迎的微积分 >

inverselaplace 1/((s+1)(s^2+2s+10))

解答

的拉普拉斯逆变换

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

解答

\frac{1}{9} e^{- t} - \frac{1}{9} e^{- t} cos (3 t)

求解步骤

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

将

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

用部份分式展开:

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

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

展开

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

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

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

L^{- 1} {\frac{1}{9 ( s + 1 )}} : \frac{1}{9} e^{- t}

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

= \frac{1}{9} e^{- t} + \frac{1}{9} (- e^{- t} cos (3 t))

整理

\frac{1}{9} e^{- t} + \frac{1}{9} (- e^{- t} cos (3 t)) : \frac{1}{9} e^{- t} - \frac{1}{9} e^{- t} cos (3 t)

= \frac{1}{9} e^{- t} - \frac{1}{9} e^{- t} cos (3 t)

流行的例子

derivative of (9x^2+4/(x^2+1))

\frac{d}{d x} (\frac{9 x ^{2} + 4}{x ^{2} + 1})

integral of ((x^2+3))/(x^2-3x+2)

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

integral of sqrt(x(3x-2))

\int x (3 x - 2) d x

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

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

integral of (2x^3+2x^2+2x+1)/(x^2+x)

\int \frac{2 x ^{3} + 2 x ^{2} + 2 x + 1}{x ^{2} + x} d x