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Differential Equation Roadblock

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the solution of the differential equation yy'(1+x2)-x(1+y2)=0 if y(0)=Sqrt(3)

    3. The attempt at a solution

    I let y=Sqrt(3) and x=0 and solve for y'.
    Integrate both sides.
    y=x +c
    Plug my points in to find C
    Sqrt(3)=0 +c
    My solution is y=x+Sqrt(3).

    Plug x+Sqrt(3) in for y, and 1 for y' into the DE.
    End up with Sqrt(3)=0, so somewhere I went wrong.
    Any help appreciated, Thanks.
  2. jcsd
  3. Oct 19, 2008 #2


    Staff: Mentor

    You are confusing y(0) with y(x). y(0) represents the y-value when x = 0. y(x) represents the y-value for an arbitrary x-value. If you were to plot the graph of y as a function of x, all you know is that the graph goes through (0, sqrt(3)).

    In your attempt at a solution, you said:
    How did you get y' = 0? This would have been correct if y(x) = sqrt(3). IOW if the graph of y were a horizontal line sqrt(3) units above the x-axis.

    If it were true that y' = 0 (it actually isn't true, though), how did you get y = x + c? If you integrate 0 with respect to x, you don't get x + C.

    The DE you are given is a separable DE. That means it is possible to get all of the things that involve y (including dy) on one side, and all of the things that involve x (and dx) on the other side. That's the approach you should take with this problem.
  4. Oct 19, 2008 #3
    Thanks for your help.

    I got it now, y=Sqrt( 4x2+3 )

    I don't know why I thought the integral of 0 with respect to X was x :rolleyes:
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