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Find the limit of a given IVP

  1. Feb 7, 2015 #1
    1. The problem statement, all variables and given/known data
    http://s14.postimg.org/an6f4t2ht/Untitled.png Untitled.png

    2. Relevant equations


    3. The attempt at a solution
    I'm not sure what they want me to do on the last part. I tried some googling and looking in my textbooks but I didn't find any examples.
    It seems to me like the function goes to infinity if it's unstable as x -> inf.

    Can somebody explain to me what exactly they want me to do?
     
  2. jcsd
  3. Feb 7, 2015 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    My suggestion would be that you actually solve the equation!
    It is just [tex]\frac{dy}{y(y- 2)(y- 4)}= dx[/tex] so integrate both sides, using "Partial Fractions" on the left.
     
  4. Feb 7, 2015 #3
    You have to consider the y value which the initial point passes through. In this case it passes through y = 3.

    Once you figure this, see which asymptotes it lies between. Do you think it will approach the stable or unstable one?

    The x-->Infinity is really just saying that a solution running through the point will go towards one of its asymptotes. It took me awhile to understand this back when I took DEs but the focus is on the y-value of the condition and y-value of asymptotes, not so much the x value

    These implicit questions want you to understand the end behavior of the derivative without necessary knowing the explicit form of y(x)
     
  5. Feb 8, 2015 #4
    I'm not sure I still really understand it, but at least I know what to do. Thanks!
     
  6. Feb 8, 2015 #5
    image003.gif
    If you look at the picture, there are different solution curves based on the points which the curve passes through (IVP). As x goes to infinity, the curves will go toward or away an equilibrium point. In this case equilibrium points are y = 2, - 1, - 2. They are called equilibrium points because if the initial value y(x) = one of the equilibrium points, the solution will go toward that asymptote even if it is an unstable asymptote (these are straight lines going through each curve).

    Try answering these:
    1. Which are stable, unstable, and semistable?
    2. What will y approach as x - - >
    infinity and y(0) = 2
    3. What will y approach as x - - > infinity and y(4) = 0
     
    Last edited: Feb 8, 2015
  7. Feb 8, 2015 #6

    epenguin

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    Homework Helper
    Gold Member

    IMHO :olduhh: Always nice when you can, but isn't it more important and more often applicable and useful to be able to give these answers without solving the equation, because most times you can't?

    I don't think the question/questioners were looking for the solution; the student could add it as afterthought to his answer to show off. :biggrin:
     
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