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(y')^2+y^2=-2 why this equation has no general solution ?

  1. Oct 15, 2010 #1

    why this differential equation has no general solution ?
  2. jcsd
  3. Oct 15, 2010 #2


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    It's non-linear. Having a general solution for these types of equations is the exception, not the rule.

    Not that there isn't a general solution because I don't know. It's just that non-linear equations rarely have general solutions
  4. Oct 15, 2010 #3
    Because the lhs is necessarily non-negative (sum of squares), whereas the lhs is negative. You can have a solution in complex numbers.
  5. Oct 16, 2010 #4
    Why not just solve it the regular way:

    [tex]\frac{dy}{\sqrt{-2-y^2}}=\pm dx[/tex]

    or [itex]y=\pm i\sqrt{2}[/tex] are solutions, maybe singular ones. Not sure. Otherwise:

    [tex]\frac{y\sqrt{-2-y^2}}{2+y^2}=\tan(c\pm x)[/tex]

    [tex]y(x)=\pm \frac{\sqrt{2}\tan(c\pm x)}{\sqrt{-\sec^2(c\pm x)}}[/tex]

    so that the solution is in the form of y(z)=u+iv
  6. Oct 17, 2010 #5


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    There is no general solution in terms of real valued functions because if y' and y are both real numbers (for a given x) then [itex](y')^2+ y^2[/itex] cannot be negative!

    Oops! Dickfore had already said that, hadn't he?
    Last edited: Oct 18, 2010
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