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DE with an initial condition

  1. Dec 11, 2006 #1
    I am asked to solve this DE with the initial condition of y(1) = 1.

    [tex](x+y)^2dx + (2xy + x^2-1)dy = 0[/tex]

    So, after working the problem out, I came to this as an answer:
    [tex]F(x,y)=\frac{1}{3}x^3 + x^2y + xy^2-y[/tex]

    My question is what do I do with the initial condition. I assume that I am just suppossed to plug 1 in somewhere, but the syntax of the initial condition does not seem very intuitive to me. What does y(1) = 1 mean?

    Thank you
  2. jcsd
  3. Dec 11, 2006 #2


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    They hint towards a x=x(y) solution.

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