1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
  7. Jul 16, 2009 #6


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook