What is inverse oflaplace 1/(2s+4) ?

The answer to inverse oflaplace 1/(2s+4) is 1/2 e^{-2t}

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inverse oflaplace 1/(2s+4)

inverse laplace

\frac{1}{2 s + 4}

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

Solution steps

L^{- 1} {\frac{1}{2 s + 4}}

= 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}

f (x) = x^{2} - 3, g (x) = 5 - x^{2}

f (x) = 6 + 4 x

e^{x + 4} + 8

\frac{x - y}{x + y}

\int \frac{y}{1 - y ^{2}} d y

What is inverse oflaplace 1/(2s+4) ?
The answer to inverse oflaplace 1/(2s+4) is 1/2 e^{-2t}