# Find the inverse Laplace transform of the given function.F(s)=\frac{2e^{-2s}}{s^2-4}

Find the inverse Laplace transform of the given function.
$F\left(s\right)=\frac{2{e}^{-2s}}{{s}^{2}-4}$
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$F\left(s\right)=\frac{2{e}^{-2s}}{{s}^{2}-4}$
$\frac{2}{\left(s-2\right)\left(s+2\right)}=\frac{A}{s-2}+\frac{B}{s+2}$
$2=A\left(s+2\right)+B\left(s-2\right)$
$A+B=0$
$2A-2B=2$
$3A=2$
$A=\frac{2}{3},B=\frac{-2}{3}$
$F\left(s\right)=\left(\frac{2{e}^{-2s}}{3\left(s-2\right)}-\frac{2{e}^{-2s}}{3\left(s+2\right)}\right)$
By applying Inverse Laplace Transform
$f\left(t\right)=\frac{2}{3}{e}^{2\left(t-2\right)}u\left(t-2\right)-\frac{2}{3}{e}^{-2\left(t-2\right)}u\left(t-2\right)$