Solve

$y^{\text{'}\text{'}}-4y^{\text{'}}+3y=\{\begin{array}{cc}1,& 0t<mn>8</mn>\\ 0,& t8\\ y(0)=y^{\text{'}}(0)=0\end{array}$

Recall that $u(t-a)$ is the unit step function shifted to the right a units.

$y(t)=f(t)u(t)-g(t)u(t-8),$ where

$f(t)=$

$g(t)=$