inverselaplace 2/(((s+2)(s^2-4)))

解答

的拉普拉斯逆变换

\frac{2}{( ( s + 2 ) ( s ^{2} - 4 ) )}

- \frac{1}{8} e^{- 2 t} - \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

求解步骤

L^{- 1} {\frac{2}{( ( s + 2 ) ( s ^{2} - 4 ) )}}

将

\frac{2}{( s + 2 ) ( s ^{2} - 4 )}

用部份分式展开:

- \frac{1}{8 ( s + 2 )} - \frac{1}{2 ( s + 2 ) ^{2}} + \frac{1}{8 ( s - 2 )}

= L^{- 1} {- \frac{1}{8 ( s + 2 )} - \frac{1}{2 ( s + 2 ) ^{2}} + \frac{1}{8 ( 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}{8 ( s + 2 )}} - L^{- 1} {\frac{1}{2 ( s + 2 ) ^{2}}} + L^{- 1} {\frac{1}{8 ( s - 2 )}}

L^{- 1} {\frac{1}{8 ( s + 2 )}} : \frac{1}{8} e^{- 2 t}

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

L^{- 1} {\frac{1}{8 ( s - 2 )}} : \frac{1}{8} e^{2 t}

= - \frac{1}{8} e^{- 2 t} - \frac{e ^{- 2 t} t}{2} + \frac{1}{8} e^{2 t}

\int \frac{2}{- x ^{2} + 4 x} d x

\frac{d}{d θ} (4 + 4 sin (θ))

\frac{d}{d x} (x - 3 x^{\frac{1}{3}})

\frac{d}{d x} (cos (3 x) \cdot 3)

d er i v a t i v e f (x) = \frac{x ^{3}}{3}