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Differential Equation Exact Solution

  1. Feb 9, 2008 #1
    1. The problem statement, all variables and given/known data
    (2x - y) + (2y-x) dy/dx = 0
    y(1) = 3
    Solve and determine where the solution is approximately valid.


    2. Relevant equations



    3. The attempt at a solution

    Ux = 2x - y
    Uy = 2y - x

    U = x^2 - yx + h(y)
    Uy = 2y -x = -x + h'(y)
    h'(y) = 2y
    h(y) = y^2

    x^2 - yx + y^2 = c
    1 - 3 + 3^2 = c = 7

    I'm not really sure where to go from here,
    x^2 - yx + y^2 = 7
    However, the answer is quite a bit different:
    y = x + sqrt( 28 - 3x^2)/2 and I can't quite see the step toward that. (x-y)^2 = 7-xy but I still can't seem to solve for y
     
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2

    quasar987

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    I have only read the last line of your post, so I don't know if your steps are right, but regarding solving for y in (x-y)^2 = 7-xy, notice that solving this equation for y(x) is the same as finding the roots of the quadratic equation y^2 + (-x)y + (x^2-7) = 0.
     
  4. Feb 9, 2008 #3

    HallsofIvy

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    And, in this case "b2- 4ac"= (-x)2- 4(1)(x[sup[2]- 7)= x2- 4x2+ 28= 28- 3x2. That's where that came from.
     
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