inverselaplace 1/(x(x^2+2x+x))

解答

的拉普拉斯逆变换

\frac{1}{x ( x ^{2} + 2 x + x )}

- \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

求解步骤

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

将

\frac{1}{x ( x ^{2} + 2 x + x )}

用部份分式展开:

- \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )}

= L^{- 1} {- \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )}}

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}{9 x}} + L^{- 1} {\frac{1}{3 x ^{2}}} + L^{- 1} {\frac{1}{9 ( x + 3 )}}

L^{- 1} {\frac{1}{9 x}} : \frac{1}{9} H (t)

L^{- 1} {\frac{1}{3 x ^{2}}} : \frac{t}{3}

L^{- 1} {\frac{1}{9 ( x + 3 )}} : \frac{1}{9} e^{- 3 t}

= - \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

\frac{d}{d x} (\frac{1}{6 x + 5})

\frac{\partial}{\partial y} (7 ln (2 x^{5} - 3 y))

a re a \frac{1}{10 x ^{2}}, x, \frac{x}{10}

\frac{d y}{d x} = k y

\frac{\partial}{\partial x} (e^{x} \cdot cos (x))