What is inverse oflaplace 1/(s^2-3) ?

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

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inverse oflaplace 1/(s^2-3)

inverse laplace

\frac{1}{s ^{2} - 3}

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

Solution steps

L^{- 1} {\frac{1}{s ^{2} - 3}}

Take the partial fraction of

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

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

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

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

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

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

\int \frac{2 x}{( 1 - x ^{2} )} d x

y^{^{'}} = \frac{y}{x} + x e^{x}

\int \frac{2}{x ^{2} ( x ^{2} + 25 )} d x

\frac{\partial}{\partial y} (6 x y - 6 y)

\frac{\partial}{\partial y} (2 x y^{3} - 1)

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