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

解答

的拉普拉斯逆变换

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

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

求解步骤

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

将

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

用部份分式展开:

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

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

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

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}

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

t an g e n t f (x) = x^{3} - 2 x + 7, at x = - 2

d er i v a t i v e x^{4} - 2 x^{2} + 5

n = 1 \sum \infty 8 e^{- 6 n}

\frac{y ^{^{'}} - e ^{- t} + 4}{y} = - 4

\int sec (x) tan (x) - 5 e^{x} d x