inverselaplace (2e^{-2s})/(s(s+2)(s+3)^2)

解答

的拉普拉斯逆变换

\frac{2 e ^{- 2 s}}{s ( s + 2 ) ( s + 3 ) ^{2}}

H (t - 2) (\frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

求解步骤

L^{- 1} {\frac{2 e ^{- 2 s}}{s ( s + 2 ) ( s + 3 ) ^{2}}}

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

其中

H (t)

是赫维赛德阶跃函数

= H (t - 2) L^{- 1} {\frac{2}{s ( s + 2 ) ( s + 3 ) ^{2}}} (t - 2)

L^{- 1} {\frac{2}{s ( s + 2 ) ( s + 3 ) ^{2}}} : \frac{1}{9} H (t) - e^{- 2 t} + \frac{8}{9} e^{- 3 t} + \frac{2 e ^{- 3 t} t}{3}

= H (t - 2) (\frac{1}{9} H (t - 2) - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

H (t - 2) = {01 t < 2 t \geq 2

= {0 \frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3} t < 2 t \geq 2

= H (t - 2) (\frac{1}{9} - e^{- 2 (t - 2)} + \frac{8}{9} e^{- 3 (t - 2)} + \frac{2 e ^{- 3 (t - 2)} ( t - 2 )}{3})

x \to a lim (2 f (x) - 3 g (x))

\frac{d}{d x} ((x^{2} + a^{2})^{e^{x^{2}}})

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

\frac{d}{d x} (lo g_{2} (3 x + 1))

x y^{^{'}} + y = x