What is inverse oflaplace s^{-3/2} ?

The answer to inverse oflaplace s^{-3/2} is (2sqrt(t))/(sqrt(pi))

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inverse oflaplace s^{-3/2}

inverse laplace

s^{- \frac{3}{2}}

\frac{2 t}{π}

Solution steps

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

Expand

s^{- \frac{3}{2}} : \frac{1}{s ^{\frac{3}{2}}}

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

= L^{- 1} {\frac{2}{π} \cdot \frac{( 2 \cdot 1 - 1 ) !! π}{2 ^{1} s ^{1 + \frac{1}{2}}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{2}{π} L^{- 1} {\frac{( 2 \cdot 1 - 1 ) !! π}{2 ^{1} s ^{1 + \frac{1}{2}}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{( 2 n - 1 ) !! π}{2 ^{n} s ^{n + \frac{1}{2}}}} = t^{n - \frac{1}{2}}

L^{- 1} {\frac{( 2 \cdot 1 - 1 ) !! π}{2 ^{1} s ^{1 + \frac{1}{2}}}} = t

= \frac{2}{π} t

= \frac{2 t}{π}

What is inverse oflaplace s^{-3/2} ?
The answer to inverse oflaplace s^{-3/2} is (2sqrt(t))/(sqrt(pi))