What is inverse oflaplace 1/(2(s+1)) ?

The answer to inverse oflaplace 1/(2(s+1)) is 1/2 e^{-t}

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inverse oflaplace 1/(2(s+1))

inverse laplace

\frac{1}{2 ( s + 1 )}

\frac{1}{2} e^{- t}

Solution steps

L^{- 1} {\frac{1}{2 ( s + 1 )}}

= L^{- 1} {\frac{1}{2} \cdot \frac{1}{s + 1}}

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{2} e^{- t}

(2.8) (0.6)

f (x) = \frac{x}{8 x ^{3} - 2 x + 9}

\frac{d}{d x} (\frac{arctan ( x )}{x})

cos (e^{x^{5} s i n (x)})

y^{^{''}} - 4 y^{^{'}} + 4 y = (18 x - 4) e^{2 x}

What is inverse oflaplace 1/(2(s+1)) ?
The answer to inverse oflaplace 1/(2(s+1)) is 1/2 e^{-t}