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Differential Equations: Find all solutions of these equations

  1. Jan 21, 2010 #1
    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.
     
  2. jcsd
  3. Jan 21, 2010 #2
    All three of them can be solved by separation of variables. That is if we have a diff. eq of the form:

    [tex]\frac{dy}{dt}=H(y,t)[/tex]

    we say it is a separable diff. eq. if [tex]H(y,t)=\frac{f(t)}{g(y)}[/tex].

    Try to do simmilar stuff with your particular functions. You've got the first step right for the 1st: moving ycos(t) to the other side(be careful of the sign though). what next? also good job with the 2nd.
     
  4. Jan 21, 2010 #3
    Oh okay. There's going to be more than one solution for all three of those right? My teacher likes to trick us like that. The second one i haven't solved yet. The fourth one is a bit confusing. I need some help with that.
     
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