What is inverse oflaplace s/(s^2-4) ?

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

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inverse oflaplace s/(s^2-4)

inverse laplace

\frac{s}{s ^{2} - 4}

\frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

Solution steps

L^{- 1} {\frac{s}{s ^{2} - 4}}

Take the partial fraction of

\frac{s}{s ^{2} - 4} : \frac{1}{2 ( s + 2 )} + \frac{1}{2 ( s - 2 )}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

L^{- 1} {\frac{1}{2 ( s - 2 )}} : \frac{1}{2} e^{2 t}

= \frac{1}{2} e^{- 2 t} + \frac{1}{2} e^{2 t}

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

\frac{2 x ^{2} + 1}{x}

\frac{d}{d x} (\frac{25 - x ^{2}}{32 x})

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

\int_{- 1}^{1} x^{2} e^{x^{3} + 1} d x

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