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

The answer to inverse oflaplace 1/((s^2-s-2)) is -1/3 e^{-t}+1/3 e^{2t}

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

inverse laplace

\frac{1}{( s ^{2} - s - 2 )}

- \frac{1}{3} e^{- t} + \frac{1}{3} e^{2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{3} e^{- t} + \frac{1}{3} e^{2 t}

x^{2} - 19, 18 x

\frac{\partial}{\partial x} (1 + x \cdot ln (x y - 5))

y^{^{'}} + x^{2} y = x^{2}

3 (cos (x^{2}) - 2 x^{2} sin (x^{2}))

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

What is inverse oflaplace 1/((s^2-s-2)) ?
The answer to inverse oflaplace 1/((s^2-s-2)) is -1/3 e^{-t}+1/3 e^{2t}