What is inverse oflaplace 1/(s^2-4) ?

The answer to inverse oflaplace 1/(s^2-4) is -1/4 e^{-2t}+1/4 e^{2t}

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inverse oflaplace 1/(s^2-4)

inverse laplace

\frac{1}{s ^{2} - 4}

- \frac{1}{4} e^{- 2 t} + \frac{1}{4} e^{2 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} - 4}}

Take the partial fraction of

\frac{1}{s ^{2} - 4} : - \frac{1}{4 ( s + 2 )} + \frac{1}{4 ( s - 2 )}

= L^{- 1} {- \frac{1}{4 ( s + 2 )} + \frac{1}{4 ( 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}{4 ( s + 2 )}} + L^{- 1} {\frac{1}{4 ( s - 2 )}}

L^{- 1} {\frac{1}{4 ( s + 2 )}} : \frac{1}{4} e^{- 2 t}

L^{- 1} {\frac{1}{4 ( s - 2 )}} : \frac{1}{4} e^{2 t}

= - \frac{1}{4} e^{- 2 t} + \frac{1}{4} e^{2 t}

(x sin (2 π x))^{^{'}}

\int_{0}^{1} π (x - x^{4})^{2} d x

e^{t + 7}

\int 5 x 1 - x^{4} d x

x (x^{2} + 4)^{5}

What is inverse oflaplace 1/(s^2-4) ?
The answer to inverse oflaplace 1/(s^2-4) is -1/4 e^{-2t}+1/4 e^{2t}