Why a explicit ODE is explicited with y of higher grade?

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Discussion Overview

The discussion revolves around the nature of explicit ordinary differential equations (ODEs) and why they are typically expressed with the highest derivative, y'', on one side of the equation. Participants explore the implications of converting implicit ODEs into explicit forms, focusing on the conventions and practicalities involved in this process.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question why an implicit ODE of the form F(x, y(x), y'(x), y''(x)) = 0 is expressed explicitly as y''(x) = f(x, y(x), y'(x)), suggesting that this transformation is not straightforward.
  • One participant provides an example, y'' - 2y' + 2y = 0, to illustrate how y'' can be expressed explicitly as y'' = 2y' - 2y, but notes that this does not fully answer the initial question.
  • Another participant suggests that the preference for explicit forms is partly due to convention and convenience, indicating that explicit forms are generally easier to work with.
  • It is mentioned that solving general implicit equations of the form F(t, y, y', ..., y^(n)) = 0 can be complex, while expressing the highest derivative explicitly as y^(n) = f(t, y, ..., y^(n-1)) is more manageable and has more established techniques.

Areas of Agreement / Disagreement

Participants express differing views on the reasons for preferring explicit forms of ODEs, with some emphasizing convention and convenience, while others focus on the complexity of implicit forms. The discussion does not reach a consensus on the underlying reasons for this preference.

Contextual Notes

Participants acknowledge that the transformation from implicit to explicit forms may depend on specific conventions and the context of the equations being solved, but do not resolve the complexities involved in this process.

Jhenrique
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Given a implicit ODE like F(x, y(x), y'(x), y''(x)) = 0, why your explicit form is y''(x) = f(x, y(x), y'(x))? Why a ODE is explicited always with y of higher grade?
 
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Jhenrique said:
Given a implicit ODE like F(x, y(x), y'(x), y''(x)) = 0, why your explicit form is y''(x) = f(x, y(x), y'(x))? Why a ODE is explicited always with y of higher grade?
What they are saying is that starting with an equation F( ... ) = 0 that involves x, y(x), y'(x), and y''(x), where y''(x) is given implicitly, a new equation can be written that gives y''(x) explicitly as a function of x and the lower-order derivatives.

A very simple example would be y'' - 2y' + 2y = 0. Here the left side is F(x, y, y', y'').
With y'' given explicitly, we have y'' = 2y' - 2y. Here the right side is f(x, y, y').
 
Mark44 said:
What they are saying is that starting with an equation F( ... ) = 0 that involves x, y(x), y'(x), and y''(x), where y''(x) is given implicitly, a new equation can be written that gives y''(x) explicitly as a function of x and the lower-order derivatives.

A very simple example would be y'' - 2y' + 2y = 0. Here the left side is F(x, y, y', y'').
With y'' given explicitly, we have y'' = 2y' - 2y. Here the right side is f(x, y, y').

?

You gave an example for my question, but not an answer...
 
AFAIK, this is partly convention and partly convenience.
 
Right, it's to make things simpler. Solving a general equation of the type

[tex]F(t,y,y^\prime,...,y^{(n)}) = 0[/tex]

is pretty horrible and there are very little techniques known. On the other hand,

[tex]y^{(n)} = f(t,y,...,y^{(n-1)})[/tex]

is way more manageable and more results about these equations are known. Also, most equations in practice show up in this form.
 

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