$\displaystyle L\left[\ \frac{d^2y}{{dt}^2}-2\frac{dy}{dt}-8y\right]=L\left[4\right] \\[2ex] \displaystyle \therefore{}L\left[\frac{d^2y}{{dt}^2}\right]-2L\left[\frac{dy}{dt}\right]-8L\left[y\right]=4\times{}\frac{1}{s} \\[2ex] \displaystyle \therefore{}\left[s^2\bar{y}-sy\left(0\right)-y^{'}(0)\right]-2\left[s\bar{y}-y(0)\right]-8\bar{y}=\frac{4}{s} \\[2ex] \displaystyle \therefore{}s^2\bar{y}-s\left(0\right)-1-2s\bar{y}-2(0)-8\bar{y}=\frac{4}{s} \\[2ex] \displaystyle \therefore{}\bar{y}\left(s^2-2s-8\right)=\frac{4}{s}+1 \\[2ex] \displaystyle \therefore{}\bar{y}\left(s-4\right)\left(s+2\right)=\frac{4+s}{s} \\[2ex] \displaystyle \therefore{}\bar{y}=\frac{(s+4)}{s\left(s-4\right)(s+2)} \\[2ex] \displaystyle \therefore{}y=L^{-1}\left[\frac{(s+4)}{s\left(s-4\right)(s+2)}\right] \\[2ex] $

$\displaystyle \therefore{}y=L^{-1}\left[\frac{-\frac{1}{2}}{s}+\frac{\frac{1}{3}}{s-4}+\frac{\frac{1}{6}}{s+2}\right](By\ Partial\ fraction\ method) \\[2ex] $

$ \displaystyle =-\frac{1}{2}L^{-1}\left[\frac{1}{s}\right]+\frac{1}{3}L^{-1}\left[\frac{1}{s-4}\right]+\frac{1}{6}L^{-1}\left[\frac{1}{s+2}\right] \\[2ex] \displaystyle =-\frac{1}{2}(1)+\ \frac{1}{3}e^{4t}+\frac{1}{6}e^{-2t} \\[2ex] \displaystyle =\frac{1}{6}(-3+2e^{4t}+e^{-2t}) \\[2ex]$