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

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

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inverse oflaplace 2/((s+2))

inverse laplace

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

2 e^{- 2 t}

Solution steps

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

= L^{- 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)}

= 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}

= 2 e^{- 2 t}

ln (\frac{( 8 x - 9 ) ^{2}}{- x - 1})

x^{\frac{11}{2}}

x^{^{'}} = - x^{2}

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

\int \frac{x + 2}{x ^{3}} d x

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