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Homework Help: One solution ODE problem

  1. Dec 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Given This ODE:

    y' = (y-2) (x^2+y)^5

    A. Show that this problem has one solution that is defined in an open segment that contains 0.

    B. Let y(x) be a solution for this problem. Prove that y(x)>2 for every x in I and conclude that y'(x)>0 in I.
    Hint: You can use the solution of the problem: y'=(y-2)(x^2+y)^5 , y(x0)=2

    Help is needed !


    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 30, 2009 #2


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    Science Advisor

    Re: Ode

    This is of the form y'= f(x,y). Your f, [itex](y-2)(x^2+ y)^5[/itex], is differentiable in both variables in any region that does not include (0,0) so you can use the "existence and uniqueess" theorem.

    Further, since y(0)= 5, [itex]y'(0)= (5-2)(0^2+ 5)^5> 0[/itex] and [itex]y'(x)= 0[/itex] only where y= 2 or [itex]y= -x^2[/itex]. The latter is impossible so any max or min must be at y= 2. Since the initial value is 5, the derivative is positive there, and can become negative only at y= 2, the function is always larger than 2.
  4. Dec 30, 2009 #3
    Re: Ode

    Hey there HallsofIvy,
    There are some things I didn't understand in your answer:
    The initial value is 5 indeed. and from y=5 the function goes up. But how can we know what happened before y=5? Maybe there was a point that was less then y=2? There can be an inflection point in y=2, and then there are values less than y=2...

    How can we solve it?

    TNX in advance!
    Last edited: Dec 30, 2009
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