inverselaplace f(t)=7t+3

解答

的拉普拉斯逆变换

f (t) = 7 t + 3

7 δ' (t) + 3 δ (t)

求解步骤

L^{- 1} {7 t + 3}

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)}

= 7 L^{- 1} {t} + L^{- 1} {3}

L^{- 1} {t} : δ^{'} (t)

L^{- 1} {3} : 3 δ (t)

= 7 δ^{^{'}} (t) + 3 δ (t)

d er i v a t i v e 1500 x + 3000 (4 x + 3)^{- 1} - 1000

\int (x^{2} - 2) (x^{3} - 6 x)^{207} d x

x \to 2 lim (3 x^{2})

s l o p e y = \frac{1}{x}

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