What is inverse oflaplace 1/(s^{3/2)} ?

The answer to inverse oflaplace 1/(s^{3/2)} is (2sqrt(t))/(sqrt(pi))

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inverse oflaplace 1/(s^{3/2)}

inverse laplace

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

\frac{2 t}{π}

Solution steps

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}

6 y^{^{''}} + 31 y = 0, y (0) = 6, y^{^{'}} (0) = 2

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

y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}

\frac{d}{d x} (\frac{- 7}{x})

What is inverse oflaplace 1/(s^{3/2)} ?
The answer to inverse oflaplace 1/(s^{3/2)} is (2sqrt(t))/(sqrt(pi))