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First order IVP (sort of)

  1. Feb 24, 2009 #1
    I just took an exam and the final problem was the following:


    dy/dx = (y-3) e^(cos(xy))

    y(1)=1

    prove that y(x) < 3 for all x for all y that are defined.

    To be honest, I tried a few things but realized I didn't know where to start. I can show that the limit of the slope from neg. infinity to infinity is zero which means 3 is an asymptote but I had no idea how to show this. Ideas?
     
  2. jcsd
  3. Feb 26, 2009 #2
    I think what you would eventually want to show is that at y=3 the surface has a max. which looks quite plausible, since y'=0=>y=3, since there is no othe possibility. Now i would take a delta neigbourhood of 3 and see how does y' change sign around it.

    The other possibility would be to try to solve that differential eq. so we would retrieve our function y, and then we could play with some inequalities in there...but i don't seem to have any ideas so far, as how to solve this diff. eq.
     
    Last edited: Feb 26, 2009
  4. Feb 26, 2009 #3

    HallsofIvy

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    Note that y'(3)= 0. One thing that tells us is that y(x)= 3 for all x satisfies the differential equation dy/dx= (y- 3) ecos(xy) with initial value y(x0)= 3.

    Further, solutions to this differential equation are unique for any initial value. (Because (y- 3)ecos(xy) is continuous in both variables and differentiable with respect to y so it satisfies the conditions for the "existence and uniqueness" theorem.)

    If y(x) satisfying that differential equation were ever equal to 3 at, say, x= x0, then we would have two distinct solutions satisfying the differential equation and y(x0)= 3, contradicting the uniqueness of y(x)= 3.

    Since any solution to the differential equation is continuous and cannot cross y= 3, if y< 3 for any value of x, it must always be less than x.
     
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