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

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

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inverse oflaplace 2/(s+1)

inverse laplace

\frac{2}{s + 1}

2 e^{- t}

Solution steps

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

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

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 + 1}}

Use Inverse Laplace Transform table:

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

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

= 2 e^{- t}

\int (3 x + 4)^{\frac{5}{2}} d x

\int_{0}^{2} (2 x - x^{2}) d x

x \to 1 + lim (\frac{x - 1}{x + 1})

\frac{3 s - 1}{s ^{2} - 8 s + 21}

\int - e^{- x^{2}} d x

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