inverselaplace 1/(s(s+4))

解答

的拉普拉斯逆变换

\frac{1}{s ( s + 4 )}

\frac{1}{4} H (t) - \frac{1}{4} e^{- 4 t}

求解步骤

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

将

\frac{1}{s ( s + 4 )}

用部份分式展开:

\frac{1}{4 s} - \frac{1}{4 ( s + 4 )}

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

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

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

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 4 t}

t an g e n t (6 e^{x} cos (x)) (0.6)

d er i v a t i v e sin (\frac{1}{2} x)

\frac{\partial}{\partial x} (\frac{12 x}{6 x ^{2} + y ^{2} + 4})

im pl i c i t \frac{d y}{d x}, x^{3} + y^{3} = 4 x y + 1

\frac{d y}{d t} = \frac{1}{2} y