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}

\int (4 x + 3)^{5} d x

\frac{d}{d x} (\frac{ln ( x )}{7})

f (x) = x^{3} - 3 x

\int 3 t^{2} sin (t) d t

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{o}})

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