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

解答

的拉普拉斯逆变换

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

\frac{- e ^{\frac{t}{2}} sin ( \frac{7 t}{2} ) + 7 e ^{\frac{t}{2}} cos ( \frac{7 t}{2} )}{7}

求解步骤

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

乘开

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

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

L^{- 1} {\frac{s}{s ( s - 1 ) + 2}} : \frac{e ^{\frac{t}{2}} sin ( \frac{7 t}{2} )}{7} + e^{\frac{t}{2}} cos (\frac{7 t}{2})

L^{- 1} {\frac{1}{s ( s - 1 ) + 2}} : e^{\frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2})

= \frac{e ^{\frac{t}{2}} sin ( \frac{7 t}{2} )}{7} + e^{\frac{t}{2}} cos (\frac{7 t}{2}) - e^{\frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2})

化简

\frac{e ^{\frac{t}{2}} sin ( \frac{7 t}{2} )}{7} + e^{\frac{t}{2}} cos (\frac{7 t}{2}) - e^{\frac{t}{2}} \frac{2}{7} sin (\frac{7 t}{2}) : \frac{- e ^{\frac{t}{2}} sin ( \frac{7 t}{2} ) + 7 e ^{\frac{t}{2}} cos ( \frac{7 t}{2} )}{7}

= \frac{- e ^{\frac{t}{2}} sin ( \frac{7 t}{2} ) + 7 e ^{\frac{t}{2}} cos ( \frac{7 t}{2} )}{7}

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

\frac{\partial}{\partial y} (3 x^{2} + y^{2})

t an g e n t y = (2 x + 1)^{2}

\frac{d ^{3}}{d x ^{3}} (cos (3 x))

\int_{1}^{\infty} \frac{1}{2 ^{x}} d x