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Y`= (x^2 + y^2)/xy ODE

  1. Aug 27, 2005 #1
    for this O.D.E. :
    y`= (x^2 + y^2)/xy
    it's unseparable, so what other methods can there be taken?
     
  2. jcsd
  3. Aug 27, 2005 #2
    It's a change of variables case (actually excercise #1 in my book on this topic), the way you do this is check to see whether f(x,y) = f(tx,ty), where f(x,y) = dy/dx.
    So if you substitute tx and ty for x and y respectively, you'll see that equality holds. Then do following substitution: y = x V(x) into the DE and see what you get with applying dy/dx and some simplifications. It should reduce to separable equation.
     
  4. Aug 27, 2005 #3

    Galileo

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    This is just like the post about the other ODE. You can write it as:

    [tex]y'=x/y+y/x[/tex]

    Again, with the substitution z=y/x it becomes:

    [tex]z+xz'=1/z+z[/tex]
    or
    [tex]z'=1/(xz)[/tex]
    which is separable.
     
  5. Aug 28, 2005 #4

    GCT

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    it's homogenous, separable, use v=y/x
     
  6. Aug 30, 2005 #5
    ok, thanks!
     
  7. Sep 3, 2005 #6
    here's my work:
    y`=x/y +x/y
    suppose y=vx
    =>v=y/x
    => y`=v+xv`
    so v+xv`=1/v+v
    => vdv=dx/x
    => v^2=ln[absolute value(x)]+c`
    => (y^2)/(x^2)=ln[absolute value(x)]+c
    => y^2=2(x^2)ln[absolute value(x)]+cx^2
    now if i take the square root on both sides, there should be a positive and negative sign on the right~
    the correct answer should only have the positive sign, but how can you be sure that it should be positive?
     
  8. Sep 3, 2005 #7

    Galileo

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    Whoopsie, you've forgot a factor 1/2 on the left side.

    You can use either sign, both will be valid solutions to the ODE. This becomes clear when you plug y back into the ODE to check if it works out. Then, with the benefit of hindsight, you could foresee this, since if y is one solution, then the other is -y and it's derivative is -y'. If you write the ODE as xyy'=x^2+y^2 you can see that if y is a solution, then -y is too.

    Ofcourse, if you're given a boundary value or initial value/condition then there will be only one solution. (otherwise the problem is ill-stated).
     
  9. Sep 4, 2005 #8
    thank you very much! :)
     
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