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

解答

的拉普拉斯逆变换

\frac{s ^{2}}{s ^{2} + 1}

δ (t) - sin (t)

求解步骤

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

将

\frac{s ^{2}}{s ^{2} + 1}

用部份分式展开:

1 - \frac{1}{s ^{2} + 1}

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - sin (t)

\frac{\partial}{\partial x} (3 x^{3} - 9 x + 9 x y^{2})

\int_{1}^{2} \frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}} d x

n or ma l y = (3 x^{2} + 1)^{2}, (0, 1)

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

d er i v a t i v e f (x) = 2 x^{2} - 1