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

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

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

inverse laplace

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

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

n = 0 \sum \infty 2 (\frac{2}{3})^{n}

n = 1 \sum \infty \frac{n}{2}

5 sin^{2} (x)

\frac{d}{d t} (8 cos (t))

x^{2} cos (x) 0.1

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