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

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

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inverse oflaplace 1/(sqrt(s))

inverse laplace

\frac{1}{s}

\frac{1}{π t}

Solution steps

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

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

= \frac{1}{π} t^{\frac{- 1}{2}}

= \frac{1}{π t}

\int sec^{2} (θ) + 6 d θ

\frac{x ^{2}}{1 + x}

\frac{d}{d r} (ln (r))

f (x) = ln (6 - x^{2} + 2 x^{4})

\int t^{3} (1 + t^{4})^{3} d t

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