Solution 2.9.5.2 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) = sy(0)+ 2y(0)+ y [1] (0) (s + 1)(s ; 3) : For the given initial conditions we then have Y (s) = s +9 (s + 1)(s; 3) = A s +1 + B s ; 3 : A = (s +1)Y (s)j s=;1 = s +9 (s ; 3) s=;1 = ;2 B = (s ; 3)Y (s)j s=4 =  s +9 (s +1)  s=3 = 3 f(t)=[3e 3t ; 2e ;t ]1(t): 1