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Differential Equation

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Verify that the indicated function [tex]y = \Phi(x)[/tex] is an explicit solution of the given first-order differential equation. Give at least one interval I of definition.

    2. Relevant equations

    [tex]2y^{'} = y^{3}cos x[/tex]
    [tex]y = (1 - sin x)^{-1/2}[/tex]

    3. The attempt at a solution

    I think I did the first part right but I am not sure about the interval I of definition.

    [tex]y^{'} = -\frac{1}{2}(1 - sin x)^{-3/2}(- cos x)[/tex]

    Left hand side:
    [tex]2y^{'} = 2(-\frac{1}{2}(1 - sin x)^{-3/2}(- cos x))[/tex]
    [tex]= -(1 - sin x)^{-3/2}(- cos x)[/tex]
    [tex]= (1 - sin x)^{-3/2}(cos x)[/tex]

    Right hand side:
    [tex]y^{3}cos x = ((1 - sin x)^{-1/2})^{3}(cos x)[/tex]
    [tex]= (1 - sin x)^{-3/2}(cos x)[/tex]

    Therefore [tex]y = (1 - sin x)^{-1/2}[/tex] is a solution.

    [tex]I = (-\infty, \pi/2)[/tex] or [tex](\pi/2, \infty)[/tex] is the interval I of definition.
  2. jcsd
  3. Sep 18, 2009 #2


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    Apparently you have noted that sin(x) can't be 1. But x = pi/2 isn't the only place that happens. You only have to give an interval and it will have to be shorter.
  4. Sep 18, 2009 #3
    Something like this?

    [tex]I = (-3\pi/2, \pi/2)[/tex] or [tex](\pi/2, 2\pi/4)[/tex]
  5. Sep 18, 2009 #4


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    Yes and no. You must have a typo in the second one.
  6. Sep 18, 2009 #5
    Yes its:
    [tex]I = (-3\pi/2, \pi/2)[/tex] or [tex](\pi/2, 5\pi/2)[/tex]

    Thank You
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