Solution 2.9.5.10 Applying the Laplace transform yields s 2 Y (s) ; sy(0) ; y [1] (0) + 2[sY (s) ; y(0)]+ 10Y (s)=0: Rearranging yields (s 2 +2s + 10)Y(s) = sy(0) + 2y(0)+ y [1] (0) = s 2 y(0) + s[y [1] (0)+ 8y(0)]+ 15 (s +1; j3)(s+1+j3) : For the given initial conditions we then have Y (s) = 3 (s +1; j3)(s+1+j3) = M s +1; j3 + M  s +1+j3 : M = (s +1; j3)Y(s) j s=;1+j3 =  3 (s +1+j3)  j s=;1+j3 = ;j=2 = 0:5 6 ;=2 Then y(t)=[e ;t cos(3t ; =2)]1(t): 1