inverselaplace 4x-3

解答

的拉普拉斯逆变换

4 x - 3

4 δ' (t) - 3 δ (t)

求解步骤

L^{- 1} {4 x - 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)}

= 4 L^{- 1} {x} - L^{- 1} {3}

L^{- 1} {x} : δ^{'} (t)

L^{- 1} {3} : 3 δ (t)

= 4 δ^{^{'}} (t) - 3 δ (t)

\frac{\partial}{\partial x} (9 e^{x - y})

n = 2 \sum \infty \frac{4}{7 ^{n}}

d er i v a t i v e f (x) = a^{5} + cos^{5} (x)

t an g e n t f (x) = x^{2} - 3 x, at x = 2

t an g e n t y = \frac{6 x}{x ^{2} + 1}, (1, 1)