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

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

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

inverse laplace

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

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

Solution steps

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

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

Use Inverse Laplace Transform table:

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

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

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

x + 3

\frac{d}{d s} (\frac{3}{s ^{2} + 9})

\int arctan (x) d x

cos^{2} (t) \frac{d y}{d t} = 1

\frac{d}{d x} (\frac{3 x - 1}{x})

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