inverselaplace (s+1)/((s+6)^2(s+2))

解答

的拉普拉斯逆变换

\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )}

\frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4} - \frac{1}{16} e^{- 2 t}

求解步骤

L^{- 1} {\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )}}

将

\frac{s + 1}{( s + 6 ) ^{2} ( s + 2 )}

用部份分式展开:

\frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{2}} - \frac{1}{16 ( s + 2 )}

= L^{- 1} {\frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{2}} - \frac{1}{16 ( 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}{16 ( s + 6 )}} + L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}} - L^{- 1} {\frac{1}{16 ( s + 2 )}}

L^{- 1} {\frac{1}{16 ( s + 6 )}} : \frac{1}{16} e^{- 6 t}

L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}} : \frac{5 e ^{- 6 t} t}{4}

L^{- 1} {\frac{1}{16 ( s + 2 )}} : \frac{1}{16} e^{- 2 t}

= \frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4} - \frac{1}{16} e^{- 2 t}

ma c l a u r in \frac{8}{3 x + 3}

d er i v a t i v e f (x) = \frac{6 cos ( x )}{7 - 9 cos ( x )}

d er i v a t i v e cos (2 x^{2})

\int_{0}^{2} \frac{3 ln ( x )}{x} d x

\frac{d}{d x} (\frac{1}{4} e^{x})