inverselaplace 1/(sqrt(2s+3))

解答

的拉普拉斯逆变换

\frac{1}{2 s + 3}

\frac{e ^{- \frac{3 t}{2}}}{2 π t}

求解步骤

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

应用逆变换法则: if

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

则

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

对于

\frac{1}{2 s + 3} : a = - \frac{3}{2}, F (s) = \frac{1}{2 s}

= e^{- \frac{3}{2} t} L^{- 1} {\frac{1}{2 s}}

L^{- 1} {\frac{1}{2 s}} : \frac{1}{2 π t}

= \frac{e ^{- \frac{3 t}{2}}}{2 π t}

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