inverselaplace {1/(s^2)-(48)/(s^5)}

解答

的拉普拉斯逆变换

{\frac{1}{s ^{2}} - \frac{48}{s ^{5}}}

t - 2 t^{4}

求解步骤

L^{- 1} {\frac{1}{s ^{2}} - \frac{48}{s ^{5}}}

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}{s ^{2}}} - L^{- 1} {\frac{48}{s ^{5}}}

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

L^{- 1} {\frac{48}{s ^{5}}} : 2 t^{4}

= t - 2 t^{4}

\frac{d}{d θ} (3 cos (θ) - 4 sin (θ))

a re a y = x^{4}, x = y^{7}, x = 0, x = 1

\int_{0}^{1} arctan (x) d x

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

\frac{\partial}{\partial s} (r + s)