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Initial value problem, end of problem Q

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the I.V.P. x2(dy/dx) = (4x2-x-2)/((x+1)(y+1)) , y(1)=1


    2. Relevant equations



    3. The attempt at a solution
    So far, I got to this:
    y2/2 + y = log(x) + 2/x + 3log(x+1) + C
    I used the initial conditions to solve for C and got:
    C = -1/2 - 3log(2)
    Substituting C back in:
    y2/2 + y = log(x) + 2/x + 3log(x+1) + (-1/2 - 3log(2))
    Now my question is, is it possible to solve for y or is this as far as I can go with the problem?
     
  2. jcsd
  3. Jan 31, 2012 #2
    Well it is a quadratic, not the nicest one but you can solve it for y.
     
  4. Jan 31, 2012 #3
    How do I set it up as a quadratic? I'm getting a little lost with all of the variables
     
  5. Jan 31, 2012 #4
    [tex]y^2 + 2y -2[\frac{2}{x} + \ln(x(x+1)^3)] = 0[/tex]

    [tex]\text{Then quadratic formula } y = \frac{-2\pm\sqrt{4 - 4(1)(-2[\frac{2}{x} + \ln(x(x+1)^3)])}}{2(1)}[/tex]

    It's horrible, but it'll give you values of y.
     
    Last edited: Jan 31, 2012
  6. Jan 31, 2012 #5
    could i just complete the square to get:
    y = +/- √[2(log(x) + (2/x) + 3log(x+1) + C) + 1] - 1
    ?
     
  7. Jan 31, 2012 #6
    Oops forgot the + C, but you get the idea.

    Completing the square would probably be a lot harder, but it should work.
     
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