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First-order nonlinear ODE

  1. Dec 1, 2012 #1
    1. The problem statement, all variables and given/known data
    We have the equation:
    [tex] y'(x)^2+2 (x+1) \left(y'(x)+x\right)+2 y(x)+2 x=0 [/tex]

    2. The attempt at a solution
    None. I don't even know how to proceed with this problem, except for, of course, expansion.
    I tried the factorization method, but no luck here. I have a feeling I need to use substitution, buuut ...

    Any clue is helpful :)

    Thank you !
     
  2. jcsd
  3. Dec 2, 2012 #2

    BruceW

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    i plugged it in to wolfram alpha, and it did not come up with an analytic solution. So I think your problem cannot be nicely solved. Did your teacher give any advice for this problem? Maybe you are supposed to give an indication of the general behaviour of the solution in the x,y phase space, rather than give an exact solution?
     
  4. Dec 3, 2012 #3
    The first thing I did after I spent 1 hour trying to figure out what to do was to plug it in into WA and, yes, the same result.
    Nope, he just said : "I hope I wrote that down correctly" :D and the problem is "Solve the equation". No hints, no advices ...
     
  5. Dec 3, 2012 #4

    Mute

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    This equation is too much for WolframAlpha. Mathematica finds a couple of solutions in the form ##F_+(x,y(x)) = C_+,~F_-(x,y(x)) = C_-##, for constant C's, involving some logarithms and square roots. The functions are also complex, but it looks like that's just because Mathematica chose to write ##\sqrt{-1+y(x)}## in several places instead of ##\sqrt{1-y(x)}##.

    Unforuntately, it's not obvious to me how mathematica might have found this solution, beyond taking a square root at some point.
     
  6. Dec 3, 2012 #5
    Tried with the regular version of Mathematica, too :) The thing is we've only used change of variables so far and just for the Bernoulli equations, so there must be a trick to reduce the degree ..
     
  7. Dec 3, 2012 #6
    Ain't that a quadratic in y'? Can't use the quadratic formula?
     
  8. Dec 4, 2012 #7
    The strange thing is, this equation has a real plot: Click here
    If did not made a silly mistake somewhere along the road :D
     
    Last edited: Dec 4, 2012
  9. Dec 4, 2012 #8

    Mute

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    I don't follow your work. In one line you have ##y'(x)## terms, and in the following line you introduce ##D_1##, which involves only y(x), and your y'(x) terms have disappeared. Were you trying to complete the square in y', as jackmell suggested? If not, you should try that. If that is what you were trying to do, perhaps you should review completing the square? Or explain to us what you were trying to do?
     
  10. Dec 4, 2012 #9
    I found an error, but I will complete the solution later, cuz I have a test ATM.
     
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