inverselaplace (8s+2)/(s^2-25)

解答

的拉普拉斯逆变换

\frac{8 s + 2}{s ^{2} - 25}

\frac{19}{5} e^{- 5 t} + \frac{21}{5} e^{5 t}

求解步骤

L^{- 1} {\frac{8 s + 2}{s ^{2} - 25}}

将

\frac{8 s + 2}{s ^{2} - 25}

用部份分式展开:

\frac{19}{5 ( s + 5 )} + \frac{21}{5 ( s - 5 )}

= L^{- 1} {\frac{19}{5 ( s + 5 )} + \frac{21}{5 ( 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{19}{5 ( s + 5 )}} + L^{- 1} {\frac{21}{5 ( s - 5 )}}

L^{- 1} {\frac{19}{5 ( s + 5 )}} : \frac{19}{5} e^{- 5 t}

L^{- 1} {\frac{21}{5 ( s - 5 )}} : \frac{21}{5} e^{5 t}

= \frac{19}{5} e^{- 5 t} + \frac{21}{5} e^{5 t}

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

n = 1 \sum \infty \frac{4}{7 ^{n}} + \frac{3}{n}

(2 x y^{2} - 7) d x + (2 x^{2} y + 6) d y = 0

\frac{d}{d x} (ln (\frac{e ^{x}}{e ^{x} - 1}))

t an g e n t f (x) = x - 1, at x = 10