What is inverse oflaplace 1/(sqrt(x)) ?

The answer to inverse oflaplace 1/(sqrt(x)) is 1/(sqrt(pi)\sqrt{t)}

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inverse oflaplace 1/(sqrt(x))

inverse laplace

\frac{1}{x}

\frac{1}{π t}

Solution steps

L^{- 1} {\frac{1}{x}}

= L^{- 1} {\frac{1}{π} \cdot \frac{( 2 \cdot 0 - 1 ) !! π}{2 ^{0} x ^{0 + \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{1}{π} L^{- 1} {\frac{( 2 \cdot 0 - 1 ) !! π}{2 ^{0} x ^{0 + \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 0 - 1 ) !! π}{2 ^{0} x ^{0 + \frac{1}{2}}}} = t^{\frac{- 1}{2}}

= \frac{1}{π} t^{\frac{- 1}{2}}

= \frac{1}{π t}

f (x) = 2 x^{2} - x

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

\int (r^{3} + 1)^{2} \cdot r^{2} d r

\int \frac{x ^{4} - x ^{3}}{x ^{5}} d x

\frac{d}{d x} (\frac{x ^{3} + 9}{x})

What is inverse oflaplace 1/(sqrt(x)) ?
The answer to inverse oflaplace 1/(sqrt(x)) is 1/(sqrt(pi)\sqrt{t)}