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Solve Initial Value Problem and Determining Interval

  1. Oct 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the initial value problem and determine at least approximately where the solution is valid

    (2x-y) + (2y-x)y' = 0
    y(1) = 3

    2. Relevant equations

    3. The attempt at a solution
    I know how to solve it, and I got the correct answer, which was:

    7 = x^2 - yx + y^2
    and then applying the quadratic formula

    x+sqrt(28 - 3x^2) / (2)

    However, I have trouble determining where the solution is valid. I understand that to find it you set 28 - 3x^2 = 0, and then solve for x (which comes out to x = +/- sqrt(28/3). However, the answers in the back of the book say that the interval of validity is IxI < sqrt(28/3). Can anyone explain how this is derived? Why is it x < sqrt(28/3)?

    Also, could someone please explain to me how to find the interval of validity in general?
  2. jcsd
  3. Oct 6, 2015 #2


    Staff: Mentor

    Actually, the solution is ##y = \frac x 2 \pm \frac{\sqrt{28 - 3x^2}}{2}##
    IOW, there are two solutions for y in terms of x. Due to the initial condition, y(1) = 3, you choose the pos. square root above.
    In order for y to be real, the quantity under the radical has to be nonnegative, so you need to solve the quadratic inequality ##28 - 3x^2 \ge 0##. Solving this results in ##|x| \le \sqrt{28/3}##, which is slightly different from what you report is the book's answer.
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