inverselaplace ((s-5))/(s(s+4)^2)

解答

的拉普拉斯逆变换

\frac{( s - 5 )}{s ( s + 4 ) ^{2}}

- \frac{5}{16} H (t) + \frac{5}{16} e^{- 4 t} + \frac{9 e ^{- 4 t} t}{4}

求解步骤

L^{- 1} {\frac{( s - 5 )}{s ( s + 4 ) ^{2}}}

将

\frac{s - 5}{s ( s + 4 ) ^{2}}

用部份分式展开:

- \frac{5}{16 s} + \frac{5}{16 ( s + 4 )} + \frac{9}{4 ( s + 4 ) ^{2}}

= L^{- 1} {- \frac{5}{16 s} + \frac{5}{16 ( s + 4 )} + \frac{9}{4 ( 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{5}{16 s}} + L^{- 1} {\frac{5}{16 ( s + 4 )}} + L^{- 1} {\frac{9}{4 ( s + 4 ) ^{2}}}

L^{- 1} {\frac{5}{16 s}} : \frac{5}{16} H (t)

L^{- 1} {\frac{5}{16 ( s + 4 )}} : \frac{5}{16} e^{- 4 t}

L^{- 1} {\frac{9}{4 ( s + 4 ) ^{2}}} : \frac{9 e ^{- 4 t} t}{4}

= - \frac{5}{16} H (t) + \frac{5}{16} e^{- 4 t} + \frac{9 e ^{- 4 t} t}{4}

a re a y = x^{2} - 4 x + 3, x - y - 1 = 0

\int (2 x^{3} + 6 e^{x}) d x

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

\frac{d}{d x} (2 x^{3} + 6 x y^{2} - 3 x^{2} + 3 y^{2})

t a y l or cos (3 x), 2.5