inverselaplace 1/(s^2-3s+2)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} - 3 s + 2}

- e^{t} + e^{2 t}

求解步骤

L^{- 1} {\frac{1}{s ^{2} - 3 s + 2}}

将

\frac{1}{s ^{2} - 3 s + 2}

用部份分式展开:

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

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

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

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

= - e^{t} + e^{2 t}

d er i v a t i v e arctan (5 x^{2} - 1)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 1 + 9 x^{2}

\int 3 sec^{4} (x) d x

x \to \frac{3 π}{4} lim (- 2 sec (x))

\frac{d x}{d t} = 6 \cdot 0.3 - 5 \cdot \frac{x}{100 + t}, x (0) = 0