inverselaplace (-5)/(s(s^2-5s+4))

解答

的拉普拉斯逆变换

\frac{- 5}{s ( s ^{2} - 5 s + 4 )}

- \frac{5}{4} H (t) + \frac{5}{3} e^{t} - \frac{5}{12} e^{4 t}

求解步骤

L^{- 1} {\frac{- 5}{s ( s ^{2} - 5 s + 4 )}}

将

\frac{- 5}{s ( s ^{2} - 5 s + 4 )}

用部份分式展开:

- \frac{5}{4 s} + \frac{5}{3 ( s - 1 )} - \frac{5}{12 ( s - 4 )}

= L^{- 1} {- \frac{5}{4 s} + \frac{5}{3 ( s - 1 )} - \frac{5}{12 ( s - 4 )}}

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{5}{4 s}} + L^{- 1} {\frac{5}{3 ( s - 1 )}} - L^{- 1} {\frac{5}{12 ( s - 4 )}}

L^{- 1} {\frac{5}{4 s}} : \frac{5}{4} H (t)

L^{- 1} {\frac{5}{3 ( s - 1 )}} : \frac{5}{3} e^{t}

L^{- 1} {\frac{5}{12 ( s - 4 )}} : \frac{5}{12} e^{4 t}

= - \frac{5}{4} H (t) + \frac{5}{3} e^{t} - \frac{5}{12} e^{4 t}

d er i v a t i v e y = - 3 x - 2

x^{^{''}} + 6 x^{^{'}} + 7 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\int_{2}^{5} (4 x - 2) d x

(x)^{^{'}}

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x