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}

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

R (x) = [x^{2} (x^{2} + 6 x)]^{4}

\frac{d}{d x} (tanh (1 + e^{2 x}))

y = \frac{1}{16} (e^{8 x} + e^{- 8 x}), - 4 \leq x \leq 4

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{2} - 1})

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