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

[Jhenrique]

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]

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').

 

[Jhenrique]

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...

 

[Mark44]

AFAIK, this is partly convention and partly convenience.

 

[micromass]

Right, it's to make things simpler. Solving a general equation of the type

F(t,y,y^\prime,...,y^{(n)}) = 0

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

y^{(n)} = f(t,y,...,y^{(n-1)})

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