inverselaplace (8s)/(3(s^2-16)^2)

解答

的拉普拉斯逆变换

\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}

- \frac{e ^{- 4 t} t}{6} + \frac{e ^{4 t} t}{6}

求解步骤

L^{- 1} {\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}}

将

\frac{8 s}{3 ( s ^{2} - 16 ) ^{2}}

用部份分式展开:

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

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

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

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

= - \frac{e ^{- 4 t} t}{6} + \frac{e ^{4 t} t}{6}

3 y + x^{4} y^{^{'}} = 2 x y

s l o p e (2, 2), (- 2, - 2)

x \to \infty lim (- 3^{x})

\int \frac{- 6 x ^{2}}{3 3 - 4 x ^{3}} d x

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