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?
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.
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.
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.
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.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?
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').