inverselaplace (12s+8)/((s-1)^5)

解答

的拉普拉斯逆变换

\frac{12 s + 8}{( s - 1 ) ^{5}}

e^{t} \cdot 2 t^{3} + \frac{5 e ^{t} t ^{4}}{6}

求解步骤

L^{- 1} {\frac{12 s + 8}{( s - 1 ) ^{5}}}

将

\frac{12 s + 8}{( s - 1 ) ^{5}}

用部份分式展开:

\frac{12}{( s - 1 ) ^{4}} + \frac{20}{( s - 1 ) ^{5}}

= L^{- 1} {\frac{12}{( s - 1 ) ^{4}} + \frac{20}{( s - 1 ) ^{5}}}

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{12}{( s - 1 ) ^{4}}} + L^{- 1} {\frac{20}{( s - 1 ) ^{5}}}

L^{- 1} {\frac{12}{( s - 1 ) ^{4}}} : e^{t} \cdot 2 t^{3}

L^{- 1} {\frac{20}{( s - 1 ) ^{5}}} : \frac{5 e ^{t} t ^{4}}{6}

= e^{t} \cdot 2 t^{3} + \frac{5 e ^{t} t ^{4}}{6}

d er i v a t i v e \frac{3}{x ^{3}} - \frac{1}{x ^{4}}

s l o p e (26 \frac{2}{3}, 9), (5, 2.5)

x \to 0 lim (x^{2} cos (x))

t an g e n t f (x) = \frac{5}{x}, at x = 5

\int \frac{3 x}{x + 1} d x