inverselaplace (e^{-7s})/((s^2-25))

解答

的拉普拉斯逆变换

\frac{e ^{- 7 s}}{( s ^{2} - 25 )}

H (t - 7) \frac{1}{5} sinh (5 (t - 7))

求解步骤

L^{- 1} {\frac{e ^{- 7 s}}{( s ^{2} - 25 )}}

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 - 7) L^{- 1} {\frac{1}{s ^{2} - 25}} (t - 7)

L^{- 1} {\frac{1}{s ^{2} - 25}} : \frac{1}{5} sinh (5 t)

= H (t - 7) \frac{1}{5} sinh (5 (t - 7))

\frac{d}{d a} (cos (a) \cdot sin (a))

\int 6 sin^{2} (x co) s^{2} x d x

\frac{d}{d x} (\frac{6}{( x - 1 ) ^{2}})

t an g e n t f (x) = 6 cos (x) + 3 sin (2 x)

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 2, y^{^{'}} (0) = 2