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

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

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inverse oflaplace s/(s^2-3)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int_{0}^{3} \frac{3}{2} x x + 1 d x

\int_{2}^{8} - \frac{x}{2} d x

\int \frac{cos ( ln ( 9 t ) )}{t} d t

\frac{d}{d x} (e^{2 c o s (3 x)})

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 9

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