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Initial Value Problem

  1. Sep 5, 2007 #1
    Im having a little issue figuring out this intial value problem.

    Solve the Initial Value Problem
    y' = (3x^2)/[(3y^2)-4] where y(0)=1

    Looks like I can solve it as a seperable DE.

    dy/dx = (3x^2)/[(3y^2)-4]

    [(3y^2)-4] dy = (3x^2) dx

    Integrating both sides.....

    (y^3) - 4y = (x^3) + c

    I don't see how to get this in terms of y = (explicitly)
    to find my c...

    Am I just missing some easy algebra or did I use the worng method for this one?

  2. jcsd
  3. Sep 5, 2007 #2
    [tex]y' = \frac{3x^2}{3y^2-4}[/tex]
    [tex](3y^2-4)y' = 3x^2[/tex]
    [tex]\int (3y^2-4)y' dx = \int 3x^2 dx[/tex].
    That means,
    [tex]y^3 - 4y = x^3 + C[/tex]
    Now at [tex]x=0[/tex] it means [tex]y=1[/tex].
    Use this to find [tex]C[/tex] and complete the problem.
  4. Sep 5, 2007 #3
    It's funny sometimes how easy something can end up being... just pluging in a point.

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