Solution 2.9.5.11 Applying the Laplace transform yields s 2 Y (s) ; sy(0); y [1] (0)+ 4[sY (s); y(0)]+ 4Y (s)=0: Rearranging yields (s 2 +4s +4)Y (s)=sy(0)+ 4y(0)+ y [1] (0);; or Y (s)= sy(0)+ 4y(0)+ y [1] (0) (s +2) 2 For the given initial conditions we then have Y (s) = 2s +5 (s +2) 2 = A s +2 + B (s +2) 2 : A = d ds h (s +2) 2 Y (s) i j s=;2 = [(d=ds)(2s+5)]j s=;2 = 2: B = h (s +2) 2 Y (s) i j s=;2 = [(2s+ 5)] j s=;2 = 1 Then, y(t)=[2e ;2t + te ;2t ]1(t): 1