(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

1) y' + ycos(t) = 0

2) y' - 2yt = 0

3) y'tan(t) + ysec^{2}(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.

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# Homework Help: Differential Equations: Find all solutions of these equations

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