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

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

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inverse oflaplace (-1)/(s+2)

inverse laplace

\frac{- 1}{s + 2}

- e^{- 2 t}

Solution steps

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

= L^{- 1} {- 1 \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)}

= - 1 \cdot 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}

= - 1 \cdot e^{- 2 t}

= - e^{- 2 t}

f (x) = \frac{1}{8 - 11 e ^{x}}

\int 5 x^{3} 3 x^{2} d x

\int (1 + x^{2}) d x

f (x) = 3 x arccos (x)

\int \frac{1}{1} d x

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