$($iii$)$ If $y(t)$ is the solution of the initial value problem $y^{\text{'}\text{'}}-y=cos(t)-sin(t)$

$y(0)=0,y^{\text{'}}(0)=0,$ then $L(y(t))(s)=$

$($A$)\frac{1}{(s-1)(s^2+1)}$

$($B$)\frac{1}{(s+1)(s^2+1)}$

$($C$)\frac{1}{(s-1)(s^2-1)}$

$($D$)\frac{1}{(s+1)(s^2-1)}$