inverselaplace (e^{-pis})/(s^2+9)

解答

的拉普拉斯逆变换

\frac{e ^{- π s}}{s ^{2} + 9}

H (t - π) \frac{1}{3} sin (3 (t - π))

求解步骤

L^{- 1} {\frac{e ^{- π s}}{s ^{2} + 9}}

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 - π) L^{- 1} {\frac{1}{s ^{2} + 9}} (t - π)

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

= H (t - π) \frac{1}{3} sin (3 (t - π))

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

\int \frac{ln ( 14 x )}{x ^{5}} d x

\int \frac{1}{t ^{2} - 2 t + 1} d t

\frac{1}{x ^{2} + y ^{2}} (x d y - y d x) = 0

l a pl a ce t r an s f or m cos (2 (t - \frac{5 π}{4}))