inverselaplace 3/s+8 1/(s(s+1))

解答

的拉普拉斯逆变换

\frac{3}{s} + 8 \frac{1}{s ( s + 1 )}

11 H (t) - 8 e^{- t}

求解步骤

L^{- 1} {\frac{3}{s} + 8 \cdot \frac{1}{s ( s + 1 )}}

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)}

= 3 L^{- 1} {\frac{1}{s}} + 8 L^{- 1} {\frac{1}{s ( s + 1 )}}

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

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

= 3 H (t) + 8 (H (t) - e^{- t})

化简

3 H (t) + 8 (H (t) - e^{- t}) : 11 H (t) - 8 e^{- t}

= 11 H (t) - 8 e^{- t}

\int sin (x) - 2 cos (x) d x

\int \frac{72}{x ^{2} 4 x ^{2} + 9} d x

n = 0 \sum \infty (5 x)^{n}

ma c l a u r in \frac{1}{1 - x ^{2}}

t an g e n t y = x, (4, 2)