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Differential equation! !

  1. Dec 5, 2006 #1
    differential equation! urgent!

    1. The problem statement, all variables and given/known data
    we have [tex]f'(x) = \frac{1}{x^2 + f(x)^2}[/tex]
    and i need to show that [tex]|f(x)| \leq \frac{5 \pi}{4}[/tex] when [tex]x \geq 1[/tex]

    2. Relevant equations

    3. The attempt at a solution

    i know this has something to do with arctan, (by the looks of f'(x), it looks similar to arctan derivative) but i really don't know how to attempt this. help pls thanks!
    Last edited: Dec 5, 2006
  2. jcsd
  3. Dec 5, 2006 #2
    I dont know how to get a trigonometric funtion in this but if y=f(x), then [tex]dy/dx=f'(x)[/tex]
    This implies that [tex]y^2dy=dx/(x^2)[/tex]
    Integrating this, you get [tex]y^3/3= -1/x +c[/tex] where c is the constant of integration. Is there some other information given? By derivative of arctan, I guess you mean [tex]d(tan^-1(x))/dx =1/(1+x^2) [/tex]
  4. Dec 5, 2006 #3
    sorry, there was mistake in what was written in latex. it's fixed now.
  5. Dec 5, 2006 #4


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    I presume chaoseverlasting's response was to some incorrectly written equation which has been edited since the equation, as now given, is not separable.

    I notice that the problem does not ask you to solve the differential equation, just to show that its solution must satisfy some property. But I wonder if there shouldn't be some initial value given? If the problem were to solve
    [tex]f'(x)= \frac{1}{x^2+ f^2(x)}[/tex]
    with initial condition f(1)= 4, then since the right hand side is defined and differentiable at (1,4), the problem has a solution in some neighborhood of (1, 4) but clearly does not satisfy "[tex]|f(x)| \leq \frac{5 \pi}{4}< 4[/tex]"
    Last edited: Dec 5, 2006
  6. Dec 5, 2006 #5
    oh right... f(1) = 1 i think this is the given value.
  7. Dec 5, 2006 #6
    then does it mean now [tex]|f(x)| \leq \frac{5 \pi}{4} [/tex] ???
    how can i show this using that differential equation?
  8. Dec 5, 2006 #7
    any help????
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