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How to decide right solution

  1. Jun 13, 2008 #1
    The problem that was given:
    [tex]{\frac {d}{dx}}y \left( x \right) =2\,{\frac {x}{{x}^{2}\cos \left( y
    \right) +4\,\sin \left( 2\,y \right) }}
    [/tex] after doing some changes we get a bernuli that looks like this
    [tex]2\, \left( {\frac {d}{dy}}x \left( y \right) \right) x \left( y
    \right) =\cos \left( y \right) \left( x \left( y \right) \right) ^{
    2}+4\,\sin \left( 2\,y \right)[/tex]
    with the initial conditions that x=f(y) goes through (1,0) (1=x,0=y)
    the solution is
    [tex]\left( x \left( y \right) \right) ^{2}=-8-8\,\sin \left( y \right) +
    9\,{{\rm e}^{\sin \left( y \right) }}[/tex]
    for these conditions the boundry is Dy=(arcsin(ln(8/9)),pi/2) which includes the point in the initial conditions the only problem i have is i dont know how to choose the final solution for x(y) thus whether is it th possitive root or the negative one
     
  2. jcsd
  3. Jun 13, 2008 #2

    tiny-tim

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    Hi supercali! :smile:

    But if it goes through (1,0), with x = 1, then it must be the positive solution, mustn't it? :confused:
     
  4. Jun 13, 2008 #3
    how stupid of me how did i not notie that??????
    after solving this creepy thing i must have been too exhausted :-)
     
  5. Jun 13, 2008 #4

    tiny-tim

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    :wink: fragilistic! :wink:
     
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