Solution 2.9.5.3 Applying the Laplace transform yields s 2 Y (s);sy(0);y [1] (0)+ 2[sY (s);y(0)];8Y (s)=0: Rearranging yields (s 2 +2s;8)Y (s)=sy(0)+ 2y(0)+ y [1] (0);; or Y (s)= sy(0)+ 2y(0)+ y [1] (0) (s 2 +2s;8) For the given initial conditions we then have Y (s) = s +10 (s + 2)(s;4) = A s +2 + B s;4 : A = (s +2)Y (s) s=;2 =  s +10 (s;4)  s=;2 = ;4=3 B = (s;4)Y (s) s=4 =  s +10 (s +2)  s=4 = 7=3 f(t)=[(7=3)e 4t ;(4=3)e ;2t ]1(t): 1