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Solving for the inital value problem Differential Equation

  1. Mar 16, 2009 #1
    I believe my question is better suited in this area, instead of Homework, but I may be wrong.

    This is what I'm given:

    (9x[tex]^{2}[/tex]+y-1) dx - (4y-x) dy =0 y(1) =3

    Solve the initial value problem and determine at least where the solution is valid.

    I did solve the problem, but I end up with this:

    A.) 3x[tex]^{3}[/tex]+xy-x-2y[tex]^{2}[/tex]=2

    When my calculator solves for Y, I get the same answer as the book does. However, I'm rather stumped at how the book gets this answer:

    B.) y = [x - (24x^3+x^2-8x-16)[tex]^{1/2}[/tex]] / 4.

    Can anyone help me to get from point A to point B? Thanks.
     
  2. jcsd
  3. Mar 16, 2009 #2

    tiny-tim

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    Hi Wellesley! :smile:

    (this should have been in the homework section)

    solve 2y2 - xy + (3x3 -x - 2) = 0 :wink:
     
  4. Mar 16, 2009 #3
    I think this should be:

    [tex]2y^2-xy-(3x^3-x-2)=0[/tex]

    :smile:

    P.S. the solution is indeed the correct one, however, aren't there suppose to be two if you rewrite it as in the book?

    coomast
     
  5. Mar 16, 2009 #4

    Sorry, I'm still not able to solve the equation for y by hand. Would you be able to show how you did it? Thanks.

    I think your equation is correct, but I'm still not able to solve for y....it has been some time since I've worked with that kind of Algebra.

    There is only one equation because only one satisfies the initial condition y (1) = 3.
     
    Last edited: Mar 16, 2009
  6. Mar 16, 2009 #5

    tiny-tim

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    Hi Wellesley! Hi coomast! :smile:

    Use the standard quadratic equation formula {-b ± √(b2 - 4ac)}/2a :wink:

    But can you please check the initial condition, y = 3 at x =1, because none of the steps seem to fit that :redface:
     
  7. Mar 16, 2009 #6
    Can you solve:

    [tex]ay^2+by+c=0[/tex]

    I think it is:

    [tex]y_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

    OK, I didn't know this. :approve:

    [edit]I should read the OP better, my fault. Anyway, is that correct?
    pfff, tiny-tim is faster....[/edit]

    coomast

    [edit 2]Something is wrong I am going to recheck everything[/edit 2]
     
    Last edited: Mar 16, 2009
  8. Mar 16, 2009 #7
    For x=1 I get y=0 and y=1/2 using the OP's solution.
    Using the solution in the book gives y=0.

    Now is y(1)=3 used for determining the constant of integration? In that case there is something wrong because it is not fullfilled by the solution.

    Wellesley, can you give some info on how you solved the DE?

    coomast

    I will come back tomorrow, it's late and I have to get up soon to get to work.
     
  9. Mar 16, 2009 #8
    Sorry about that. I copied the inital condition from the problem above. It should be y (1) = 0

    Thanks guys. I can't believe the answer is that simple...Here I am thinking that I had to somehow complete a square with a cubic exponent, or worse. Instead, it was the quadratic formula that never seems to go away.
     
    Last edited: Mar 16, 2009
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