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Verifying relations

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data



    Verify that the relation x^2 + y^2 = 1 is a solution to the differential equation:
    dy/dx = xy/(x^2 - 1)

    Can anyone point me in the right direction on how to begin to solve this problem? Do I take integral of the DE and just plug into equation?
     
    Last edited: Jan 15, 2009
  2. jcsd
  3. Jan 15, 2009 #2

    rock.freak667

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    You could take x^2 + y^2 = 1 and find dy/dx and see if it is the same as the DE
     
  4. Jan 15, 2009 #3
    dy/dx is just 2x+2y. So it does not equal correct?
     
  5. Jan 15, 2009 #4

    rock.freak667

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    2x+2y(dy/dx)=0 is what you get when you differentiate implicitly with repect to x. Now just find dy/dx and do a bit of algebra.
     
  6. Jan 15, 2009 #5
    I'm sorry but I'm really slow at DE's.

    2x+2y(dy/dx)=0

    What do you mean find dy/dx. I thought dy/dx was 2x+2y=0.

    Are you saying to solve for dy/dx in the first equation?
     
  7. Jan 15, 2009 #6

    rock.freak667

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    Yes, make dy/dx the subject of the equation.

    EDIT: is the relation supposed to be y2-x2=1 ?
     
    Last edited: Jan 15, 2009
  8. Jan 15, 2009 #7
    2y(dy/dx) = -2x

    divide by 2y:

    dy/dx = -2x/2y
     
  9. Jan 15, 2009 #8

    rock.freak667

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    So dy/dx= -x/y right?

    Now multiply the numerator and denominator by y.

    (Forget my edit in the previous post)
     
  10. Jan 15, 2009 #9
    dy/dx = -xy

    That is the answer to the problem? So the DE is not a solution.
     
  11. Jan 15, 2009 #10

    rock.freak667

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    If dy/dx=-x/y and you multiply by y/y, do you really get dy/dy=-xy or -xy/y2?
     
  12. Jan 15, 2009 #11
    yeah, your right... dy/dx = -xy/y^2

    But the top y would just cancel and the y^2 would become y again so why do that?
     
  13. Jan 15, 2009 #12

    rock.freak667

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    Don't cancel out anything. Keep it the way it is. Now from the relation x2+y2=1, what is y2 equal to?
     
  14. Jan 15, 2009 #13
    y^2 = 1-x^2
     
  15. Jan 15, 2009 #14

    rock.freak667

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    Now in the equation

    [tex]\frac{dy}{dx}=\frac{-xy}{y^2}[/tex]


    Replace y2.
     
  16. Jan 15, 2009 #15
    dy/dx = -xy/1-x^2

    So they do equal. If you multiply by -1, it gives the DE.
     
  17. Jan 15, 2009 #16

    rock.freak667

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    and if it gives the DE, it is a solution of it.
     
  18. Jan 15, 2009 #17
    Cool..Thank you very much for your help and patience.
     
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