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Behavior of DE as t approaches infinity

  1. Jun 27, 2012 #1
    1. The problem statement, all variables and given/known data

    y' = -2 + t - y

    Draw a slope field and determine behavior as t -> infinity

    2. Relevant equations



    3. The attempt at a solution

    This is the first DE in my book that includes t in the differential equation. The slope field looks pretty wacky. For the others I was able to see the behavior easily through the slope field because it didn't change along the t axis. Now, just looking at the slope field won't do.

    How does one determine the behavior as t approaches infinity here? (without solving the equation!)

    Thanks!
     
  2. jcsd
  3. Jun 28, 2012 #2
    Is [itex]y[/itex] constant as [itex]t[/itex] increases? If so, then [itex]y'[/itex] increases as [itex]t[/itex] increases; furthermore, [itex]y'[/itex] increases without bound as [itex]t[/itex] increases without bound.
     
    Last edited: Jun 28, 2012
  4. Jun 28, 2012 #3

    HallsofIvy

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    Well, no, y is NOT constant as t changes! The whole point of a differential equation is that y is a function of t. And how could y' increase (or be anything but 0) if y is constant?

    1MileCrash, what does the slope field look like for large t? If it looks, say, like a straight line, try y= Ax+ B and put into the equation to find A and B.
     
  5. Jun 28, 2012 #4
    Well think in terms Of equilibrium points. We know that y is max/min/or saddle at y'=0, so your max or min would fall on y=t-2, which would imply y goes to infinity as t does
     
  6. Jun 28, 2012 #5
    How should I draw a slope field for large t? What should I do with the y's? Make them large too? Leave them the same? Couldn't the slope field be any number of things for large t depending on what y's I consider?

    Thanks again.
     
  7. Jun 30, 2012 #6
    Perhaps I shouldn't have phrased whether [itex]y[/itex] varies or not as a question. Of course [itex]y[/itex] varies; however, what can one say about the general behavior of [itex]y'[/itex] as [itex]t[/itex] increases without bound without restricting [itex]y[/itex] to a constant value?

    You ask how [itex]y'[/itex] would increase if [itex]y[/itex] were constant. [itex]y' = -2 + t - y[/itex] is given. Let [itex]y = y_0[/itex] for some constant [itex]y_0[/itex]. Then, [itex]y' = -2 + t - y_0[/itex], from which it is clear [itex]y'[/itex] is an increasing function of [itex]t[/itex].

    Alternatively, consider the slope field for the given differential equation. Restricting [itex]y[/itex] to a constant value [itex]y_0[/itex] is equivalent to restricting the slope field to only those points on the line [itex]y = y_0[/itex]. The restricted slope field shows [itex]y'[/itex] is an increasing function of [itex]t[/itex].

    In answer to your last question, yes. The slope at any given point on the slope field depends on both [itex]t[/itex] and [itex]y[/itex]. The slope field calculator hosted by Rutgers at http://www.math.rutgers.edu/~sontag/JODE/JOdeApplet.html might be useful to you.
     
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