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

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Discussion Overview

The discussion revolves around the differential equation (y')^2 + y^2 = -2, specifically examining why this equation does not have a general solution in the context of real-valued functions.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants note that the equation is non-linear, suggesting that general solutions for non-linear equations are rare.
  • One participant points out that the left-hand side (lhs) of the equation is necessarily non-negative (being a sum of squares), while the right-hand side (rhs) is negative, indicating a fundamental inconsistency for real solutions.
  • Another participant proposes a method to solve the equation, suggesting that complex solutions may exist, and provides a transformation involving complex numbers.
  • One participant emphasizes that there is no general solution in terms of real-valued functions, reiterating the point about the non-negativity of the lhs compared to the negative rhs.

Areas of Agreement / Disagreement

Participants generally agree that the equation does not have a general solution in the context of real-valued functions due to the inherent contradictions in the equation. However, there is some exploration of the possibility of complex solutions, indicating a lack of consensus on the nature of solutions.

Contextual Notes

Participants express uncertainty regarding the nature of potential solutions, particularly the distinction between real and complex solutions, and the implications of non-linearity on the existence of general solutions.

young_eng
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hi

(y')^2+y^2=-2
why this differential equation has no general solution ?
 
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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
 
Because the lhs is necessarily non-negative (sum of squares), whereas the lhs is negative. You can have a solution in complex numbers.
 
young_eng said:
hi

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

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:<br /> <br /> [tex]\frac{y\sqrt{-2-y^2}}{2+y^2}=\tan(c\pm x)[/tex]<br /> <br /> [tex]y(x)=\pm \frac{\sqrt{2}\tan(c\pm x)}{\sqrt{-\sec^2(c\pm x)}}[/tex]<br /> <br /> so that the solution is in the form of y(z)=u+iv[/itex]
 
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?
 
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