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Critical points of differential equation

  1. Jul 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the location and type of all critical points of the given equations and sketch the phase portrait

    y"+cosy=0



    3. The attempt at a solution

    I've done some like this before but they were all systems of equations. I'm actually not sure how to do the simpler version >_< I think this is asking for maxima and minima? is that right?
     
  2. jcsd
  3. Jul 10, 2012 #2

    HallsofIvy

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    It's NOT simpler. This is only one equation but is second order. What you want is two first order equations. Define x by x= y'. Then y''= x' so y''+ y= 0 becomes x'+ y= 0 or x'= -y.

    You now have the two equations x'= -y and y'= x.
     
  4. Jul 10, 2012 #3
    Okay, so for my problem I have
    x=y'
    x'=y"
    so I substitute and
    x'+cosy=0
    x'=-cosy

    so my system of equations is;

    y'=x
    x'=-cosy

    right? Then I just solve like it's a system of equations and look for the singularities as the critical points?
     
  5. Jul 10, 2012 #4
    Ah! No wait, there's more! I know what I'm doing now, silly me. I set x' and y' to zero and solve for the points.

    x'=0, ∴x=0
    y'=0, -cosy=0, ∴y=(1/2)(2n+1)pi

    So I have infinite points along the y-axis. I used

    http://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html [Broken]

    to graph the phase portrait and found that every other critical point is a saddle point, then a center point.

    SO! how do I say that one is a saddle, then the next is a center?
     
    Last edited by a moderator: May 6, 2017
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