Consider the initial value problem $y^{\text{'}\text{'}}-4y^{\text{'}}+4y=(t-2{)}^2{u}_2(t)$

$y(0)=1$ and $y^{\text{'}}(0)=2,$ and let $y(t)$ be its solution. Then

A$.y(t)=L^{-1}\{\frac{2e^{-2s}}{s^3(s-2{)}^2}+\frac{s+6}{(s-2{)}^2}\}$

$(t)$

B$.y(t)=L^{-1}\{\frac{2e^{-2s}}{s^3(s-2{)}^2}+\frac{1}{s-2}\}$

$(t)$

C$.y(t)=L^{-1}\{\frac{e^{-2s}}{s^3(s-2{)}^2}+\frac{s+6}{(s-2{)}^2}\}$

$(t)$

D$.y(t)=L^{-1}\{\frac{2e^{-2s}}{s^3(s-2{)}^2}+\frac{1}{(s-2{)}^2}\}(t)$

$(t)$

E$.y(t)=L^{-1}\{\frac{(16s^2+8s+2)e^{-2s}}{s^3(s-2{)}^2}+\frac{1}{(s-2{)}^2}\}$