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

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

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inverse oflaplace 1/(sqrt(s+1))

inverse laplace

\frac{1}{s + 1}

\frac{e ^{- t}}{π t}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

L^{- 1} {\frac{1}{s}} : \frac{1}{π t}

= \frac{e ^{- t}}{π t}

t an g e n t f (x) = 6 tan (x), at x = \frac{π}{6}

x y^{^{'}} = y + x^{2} - y^{2}

\int (1 + tan (θ))^{5} sec^{2} (θ) d θ

\frac{d y}{d x} = \frac{y ( 2 x ^{3} - y ^{3} )}{x ( 2 x ^{3} - 3 y ^{3} )}

\frac{d}{d x} (\frac{2 x}{3})

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