1. The problem statement, all variables and given/known data Let y be the solution of the initial value problem: y'' + 2y' + 2y = 0 , y(0) = 0 , y'(0) = 5 The maximum value of y over 0 less than or equal to x less than infinity is ??. 3. The attempt at a solution r = -2 +/- i I solved it: y = c1 e^-2x cos(x) + c2 e^-2x sin(x) y' = -2c1 e^-2x cos(x) - c1 e^-2x sin(x) - 2 c2 e^-2x sin(x) + c2 e^-2x cos(x) y = c1 = 0 y' = -2c1 + c2 = 5 => c2 = 5 y = 5 e^-2t sin(x) But I am terrible at trig. What is the maximum value of y? I got (x = -3pi/2 so therefore max y is 5 e^(6pi/2) but that is incorrect. I have one more attempt and don't want to lose out on this question. Thanks.