Answer
$y(t)=\frac{7}{3}e^{-2t}+\frac{11}{3}e^{4t}$
Work Step by Step
Taking the Laplace transform of both sides of the given differential equation and imposing the initial conditions yields:
$[s^2(Y)-sY(0)-y'(0)]-2[s(Y)-y(0)]-8Y(s)=\frac{5}{s}\\
s(Y)(s^2-2s-8)=\frac{5}{s}+s-2$
which implies that:
$Y(s)=\frac{5}{s^2-2s-8}+\frac{s-2}{s^2-2s-8}\\
=\frac{5}{(s-4)(s+2)}+\frac{s-2}{(s-4)(s+2)}\\
=\frac{5}{3(s+2)}+\frac{10}{3(s-4)}+\frac{1}{3(s-4)}+\frac{2}{3(s+2)}\\
=\frac{7}{3(a+2)}+\frac{11}{3(s-4)}$
Taking the inverse Laplace transform of both sides gives
$y(t)=L^{-1}[\frac{}{3(a+2)}+\frac{11}{3(s-4)}]$
We finally obtain
$y(t)=\frac{7}{3}e^{-2t}+\frac{11}{3}e^{4t}$
