Solution 2.9.5.4 For y [2] +2y [1] ;3y(t)=0;; y(0) = 0;;y [1] (0) = 4;; Applying the Laplace transform yields s 2 Y (s);sy(0);y [1] (0)+ 2[sY (s);y(0)];3Y (s)=0: Rearranging yields (s 2 +2s;3)Y (s)=sy(0)+ 2y(0)+ y [1] (0);; or Y (s)= sy(0)+ 2y(0)+ y [1] (0) (s 2 +2s;3) For the given initial conditions we then have Y (s) = 4 (s + 3)(s;1) = A s +3 + B s;1 : A = (s +3)Y (s) s=;3 =  4 (s;1)  s=;3 = ;1 B = (s;1)Y (s) s=1 =  4 (s +3)  s=1 = 1 f(t)=[e t ;e ;3t ]1(t): 1