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

The answer to inverse oflaplace 2/(s^2+s-2) is 2/3 e^t-2/3 e^{-2t}

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

inverse laplace

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

\frac{2}{3} e^{t} - \frac{2}{3} e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{2}{s ^{2} + s - 2} : \frac{2}{3 ( s - 1 )} - \frac{2}{3 ( s + 2 )}

= L^{- 1} {\frac{2}{3 ( s - 1 )} - \frac{2}{3 ( 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{2}{3 ( s - 1 )}} - L^{- 1} {\frac{2}{3 ( s + 2 )}}

L^{- 1} {\frac{2}{3 ( s - 1 )}} : \frac{2}{3} e^{t}

L^{- 1} {\frac{2}{3 ( s + 2 )}} : \frac{2}{3} e^{- 2 t}

= \frac{2}{3} e^{t} - \frac{2}{3} e^{- 2 t}

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What is inverse oflaplace 2/(s^2+s-2) ?
The answer to inverse oflaplace 2/(s^2+s-2) is 2/3 e^t-2/3 e^{-2t}