Solution 2.9.5.9 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) = 2 (s;2) 2 = A s;1 + B (s;1) 2 : A = d ds h (s;2) 2 Y (s) i s=2 = d ds (2) = 0 B = [2] s=2 = 2 y(t)=te 2t 1(t) 1