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

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

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inverse oflaplace s/(s^2-2)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

y^{^{'}} = 7 x + y

\frac{d y}{d x} + 3 x y^{2} = 0, y (0) = 1

\frac{d}{d x} ((\frac{6 x ^{2}}{4 - x})^{3})

n = 0 \sum \infty \frac{n}{n ^{4} + 1}

\frac{d}{d x} (\frac{x + 9}{x ^{3} + x - 8})

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