2.9.5.1 Applying the Laplace transform yields s 2 Y (s);sy(0);y [1] (0)+ sY (s);y(0);2Y (s)=0: Rearranging yields (s 2 + s;2)Y (s)=sy(0)+ y(0)+ y [1] (0);; or Y (s)= sy(0) + y(0)+ y [1] (0) (s 2 + s;2) For the given initial conditions we then have Y (s) = 3 (s + 2)(s;1) = A s +2 + B s;1 : A = (s +2)Y (s) s=;2 =  3(s +2) (s +2)(s;1)  s=;2 =  3 (s;1)  s=;2 = ;1 B = (s;1)Y (s) s=1 =  3(s;1) (s +2)(s;1)  s=1 =  3 (s +2)  s=1 = 1 Then f(t)=[e t ;e ;2t ]1(t): 1