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}

n = 1 \sum \infty \frac{n !}{n 5 ^{n}}

x \to 0 lim (\frac{4 ^{x} - 1}{x})

(\frac{3 x ^{2}}{2 x + 4})^{3}

2 x y^{^{'}} + y = 6 x, y (4) = 25

y = - 2 x^{3}, (1, - 2)

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