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Seperable equations

  1. Apr 28, 2006 #1
    Ok, I'm having trouble with the last part of this problem. After solving for the constant C, I get:
    y^2 - 2y = x^3 + 2x^2 +2x + 3

    My question is, how do I solve this in terms of Y? The only instruction my book gives me is "To obtain the solution explicitly we must solve for y in terms of x. This is a simple matter in this case since the equation is quadratic in y" and then they jump to the solution:
    y = 1 +/- sqrt(x^3 + 2x^2 +2x + 4)
  2. jcsd
  3. Apr 28, 2006 #2


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    Gold Member

    Well you said it yourself: it's a quadratic in terms of y.

    Do you know how to solve quadratic equations?
  4. Apr 28, 2006 #3

    So would the expression x^3 + 2x^2 +2x + 3 be treated as the 'c' value in the quadratic eq?
  5. Apr 28, 2006 #4


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    Gold Member


    Remember that the quadratic equation lets you solve for a certain variable (we'll call it w here) when you have an equation in the form of [itex]aw^2+bw+c=0[/itex]
  6. Apr 28, 2006 #5
    Alright I got it, thanks a lot
  7. Apr 29, 2006 #6


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    Staff Emeritus
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    Actually, no. Since your equation is y^2 - 2y = x^3 + 2x^2 +2x + 3
    and the quadratic is normally written ay^2+ by+ c= 0,
    c= -x^3- 2x^2- 2x- 3.
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