One of my homework problems this week was: Verify that y = sin(4t) + 2cos(4t) is a solution to the following initial value problem. 2y'' + 32y=0; y(0)=2, y'(0)=4 Find the maximum of |y(t)| for -infinity< t < infinity. Verifying that the given y equation is a solution is easy, all there is to do is derive twice, plug in the new equations, and show that 0 = 0. I have no trouble with that. My question is with the second part, about finding the maximum positive value of y(t). I would think that no matter how large t gets, sin and cos will always oscillate between -1 and 1. When cos is 1, obviously sin is 0. Putting in sin(4t) = 0 and cos(4t) = 1 gives me a maximum value of 2 for any y(t) as t approaches infinity. I graphed the function on a calculator to confirm this, just in case, and the graph appears to agree with me. My confusion is that 2 isn't the correct answer. According to the website we do our homework on, the answer should be sqrt(5). There are no questions like this worked out in the book and I can't find anything online. Can anybody explain to me why y(t) has a maximum value of sqrt(5) as t approaches infinity? Thank you!