In Exercises 31-40, find the unique solution of the second-order initial value problem. 31. "+6y+5y=0, y(0) = 0, y'(0) = 3 32. " 16y0, y(0) = 2, y'(0) = 33. " 12y0, y(0) = 0, y'(0) = 1 34. 12" = 5y2y = 0, y(0) = 1, y'(0) - 35. " 8y0, y(0)=-1, y'(0) = 2 36. y" +4y+ 4y = 0, 37. y4y4y = 0, 38. 4" 4' + y = 0, dy dy 40. 9. dy 12 dr² dx 39. 4 d²y dx +12 +9y= 0, y(0) = 2, dx + 4y = 0, y(0) = -1, y(0) = 0, y'(0) y(0) = 1, y'(0) = 0 y(0) = 4, y'(0) = 4 dy dx (0) = 1 dy dx dr (0) = = 1