1. The problem statement, all variables and given/known data 1) y' + ycos(t) = 0 2) y' - 2yt = 0 3) y'tan(t) + ysec2(t) = 0 4) Let p(x) be an arbitrary polynomial function. Note that the following is a polynomial solution to y" = p(x) + y: y = -p - p" - p(4) - p(6) - ... - p(n) where n is the smallest positive even integer such that p(n+1) is the zero polynomial. Prove that the following differential equation has a polynomial solution: y''' = p(x) + 2y 2. Relevant equations 3. The attempt at a solution 1) First, I tried moving the ycos(t) to the right, but I don't know where to go from there. y' = ycos(t) 2) I moved 2ty to the right. Integrated it. y' = 2ty 3) This one is just a beast. I was never good at trigonometry. I moved the ysec^2t over to the right and canceled out the trig functions to get down to this: y' = ysec(t) I don't know where to go from here. 4) I don't even know how to go about doing this problem.