What is inverse oflaplace 8/((s^2+1)(s^2-1)) ?

The answer to inverse oflaplace 8/((s^2+1)(s^2-1)) is -4sin(t)-2e^{-t}+2e^t

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inverse oflaplace 8/((s^2+1)(s^2-1))

inverse laplace

\frac{8}{( s ^{2} + 1 ) ( s ^{2} - 1 )}

- 4 sin (t) - 2 e^{- t} + 2 e^{t}

Solution steps

L^{- 1} {\frac{8}{( s ^{2} + 1 ) ( s ^{2} - 1 )}}

Take the partial fraction of

\frac{8}{( s ^{2} + 1 ) ( s ^{2} - 1 )} : - \frac{4}{s ^{2} + 1} - \frac{2}{s + 1} + \frac{2}{s - 1}

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

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

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

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

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

= - 4 sin (t) - 2 e^{- t} + 2 e^{t}

\int \frac{a ^{2} - x ^{2}}{x ^{4}} d x

x \to 2 lim (x + 3)

y = ln (lo g_{e} (2 x))

n = 1 \sum \infty (x - 2)^{n}

\int cos (x) 1 + sin^{2} (x) d x

What is inverse oflaplace 8/((s^2+1)(s^2-1)) ?
The answer to inverse oflaplace 8/((s^2+1)(s^2-1)) is -4sin(t)-2e^{-t}+2e^t