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1st order dif eq help

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

    (2x-y)dx+(2y-x)dy=0 y(1)=3

    solve the given initial value problem and determine at least approx where the solution is valid?


    this should be simple right? I got 2y+2x=C and then 2y+2x=8. It doesnt match the answer in the back of the book at all, and I do see how they could have gotten the answer in the back. Is my answer right? The book says y=[x+sqrt(28-3x^2)]/2, abs(x)<sqrt(28/3)? or is the book right, and if so how do you get to their answer?
     
  2. jcsd
  3. Jul 15, 2009 #2
    The book is right, this is an exact differential equation. You can tell this since the partial derivative of the term multiplied by dx, namely (2x-y), with respect to y is equal to the partial derivative of the term multiplied by dy, or (2y - x), with respect to x (since both partials are -1).

    Check out http://www.sosmath.com/diffeq/first/exact/exact.html. Follow the method step by step and that will lead you straight to the solution.
     
  4. Jul 15, 2009 #3
    thanks I got to x^2-yx+y^2=7 now. I have no clue how to seperate this right now, is there a trick, did I do something wrong.
     
  5. Jul 15, 2009 #4
    No you basically got this. To complete the problem, subtract 7 from both sides and view the equation as a quadratic in y (treating other quantities as constants) and apply the quadratic formula.
     
  6. Jul 15, 2009 #5
    And how did you solve this
    (2x-y)dx+(2y-x)dy=0
    ?
     
  7. Jul 16, 2009 #6

    HallsofIvy

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    But, generally speaking, it is not necessary to "solve" the equation for y. [itex]x^2- xy+ y^2=7[/itex] is a perfectly good solution.
     
  8. Jul 16, 2009 #7
    solved, thanks guys, cant believe I couldnt see those
     
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