Use Laplace Transform to solve

$($a$)y^{\text{'}\text{'}}+4y^{\text{'}}+6y=1+e^{-t},y(0)=0,y^{\text{'}}(0)=0.$

$($b$)y^{\text{'}\text{'}}+4y^{\text{'}}+5y=(t-2),y(0)=0,y^{\text{'}}(0)=0.$

$($c$)y^{\text{'}}(t)=cost+{\displaystyle \underset{0}{\overset{t}{}}}y()cos(t-)d,y(0)=1,$ for $y(t).$

$($d$)y^{\text{'}}-5y=f(t),$ where $f(t)=\{\begin{array}{cc}{t}^2,& 0t<mn>1</mn>\\ 0,& t1\\ y(0)=1\end{array}.$