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Math Help - Why are exact ODEs called exact?

  1. #1
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    Why are exact ODEs called exact?

    No, it's not the start of a joke!

    I'm just wondering why are exact ODEs called exact?

    All I can think of so far is that they don't end up with an integration constant on one of the variables.

    But as the solution is e.g. f(x,y) = c it's still going to represent a family of solutions.

    Any ideas? It's probably really obvious It's been annoying me for a few days now.

    Thanks and merry Christmas and new year to you all!
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  2. #2
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    I'm not sure how much maths you've done before but here goes as an explination.

    A first order ODE can be written in the form
     p(x) \frac{dy}{dx} + q(x) y = r(x)

    If it is exact the the LHS can be directly written as

     \frac{d}{dx}(s(x)y) = r(x)

    I can guess the reason they are called exact is related to the explanation given here
    Exact First-Order Ordinary Differential Equation -- from Wolfram MathWorld

    Which is obvious if you have done some multi-variable calculus

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  3. #3
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    The term is from "exact differential".

    While any thing of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz looks like the differential of some F(x,y,z), it may not be. Given any function F(x,y,z), its differential is dF(x,y,z)= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}dz. that comes from the chain rule: \frac{dF(x,y,z)}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}+ \frac{\partial F}{\partial z}\frac{dz}{dt} for any parameter, t.

    In order that f(x,y)dx+ g(x,y)dy+ h(x,y)dz be a "real" (or "exact") differential, we must have f(x,y)= \frac{\partial F}{\partial x} and g(x,y)= \frac{\partial F}{\partial y}, so that the mixed derivatives are equal: \frac{\partial^2 F}{\partial x\partial y} = \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} = \frac{\partial^2 F}{\partial y\partial x} and the same for the other mixed derivatives.

    Back in Calculus III such a thing was called an "exact differential" (some texts use the term "conservative" in analogy with "conservative force fields" in physics but I dislike using physics terms in mathematics). In the theory of "differential forms" on manifolds, a differential form, \omega, of order n, is called "exact" if there exist some differential form \alpha, of order n-1, such that \omega= d\alpha.
    Last edited by HallsofIvy; December 27th 2009 at 06:26 AM.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    The term is from "exact differential".

    While any thing of the form f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz looks like the differentiall of some F(x,y,z), it may not be. Given any function F(x,y,z), its differential is dF(x,y,z)= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}dz. that comes from the chain rule: \frac{dF(x,y,z)}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}+ \frac{\partial F}{\partial z}\frac{dz}{dt} for any parameter, t.

    In order that f(x,y)dx+ g(x,y)dy+ h(x,y)dz be a "real" (or "exact") differential, we must have and the same for the other mixed derivatives.

    Back in Calculus III such a thing was called an "exact differential" (some texts use the term "conservative" in analogy with "conservative force fields" in physics but I dislike using physics terms in mathematics). In the theory of "differential forms" on manifolds, a differential form, \omega, of order n, is called "exact" if there exist some differential form [itex]\alpha[/itex], of order n-1, such that [itex]\omega= d\alpha[/itex].
    Thanks to you both for your answers. It will be clear how much maths I have done if you keep reading!

    One crucial thing was a lack of familiarity with differentials as a thing in their own right. So I now have a broader knowledge.

    When it came to f(x,y)= \frac{\partial F}{\partial x} and g(x,y)= \frac{\partial F}{\partial y}, so that the mixed derivatives are equal: \frac{\partial^2 F}{\partial x\partial y} = \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} = \frac{\partial^2 F}{\partial y\partial x} I had to ponder on and off for two days until it suddenly appeared in my head that this is just seen by taking the partial w.r.t y of f and the partial w.r.t to x of g. Now you see how much maths I have actually done!

    Anyway, fantastic, I now understand and have learnt some extra theory - not just the mechanical approach in which exact differential equations are solved. Am very happy!

    Merry Christmas and have a good 2010!

    i
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  5. #5
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    Quote Originally Posted by iota View Post
    Thanks to you both for your answers. It will be clear how much maths I have done if you keep reading!

    One crucial thing was a lack of familiarity with differentials as a thing in their own right. So I now have a broader knowledge.

    When it came to f(x,y)= \frac{\partial F}{\partial x} and g(x,y)= \frac{\partial F}{\partial y}, so that the mixed derivatives are equal:
    Mixed derivatives are equal as long as they are continuous.

    \frac{\partial^2 F}{\partial x\partial y} = \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} = \frac{\partial^2 F}{\partial y\partial x} I had to ponder on and off for two days until it suddenly appeared in my head that this is just seen by taking the partial w.r.t y of f and the partial w.r.t to x of g. Now you see how much maths I have actually done!

    Anyway, fantastic, I now understand and have learnt some extra theory - not just the mechanical approach in which exact differential equations are solved. Am very happy!

    Merry Christmas and have a good 2010!

    i
    Follow Math Help Forum on Facebook and Google+

  6. #6
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    Quote Originally Posted by iota View Post
    Thanks to you both for your answers. It will be clear how much maths I have done if you keep reading!

    One crucial thing was a lack of familiarity with differentials as a thing in their own right. So I now have a broader knowledge.

    When it came to f(x,y)= \frac{\partial F}{\partial x} and g(x,y)= \frac{\partial F}{\partial y}, so that the mixed derivatives are equal:
    Mixed derivatives are equal as long as they are continuous. If they are not continuous, they are not necessarily equal.

    \frac{\partial^2 F}{\partial x\partial y} = \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} = \frac{\partial^2 F}{\partial y\partial x} I had to ponder on and off for two days until it suddenly appeared in my head that this is just seen by taking the partial w.r.t y of f and the partial w.r.t to x of g. Now you see how much maths I have actually done!

    Anyway, fantastic, I now understand and have learnt some extra theory - not just the mechanical approach in which exact differential equations are solved. Am very happy!

    Merry Christmas and have a good 2010!

    i
    Follow Math Help Forum on Facebook and Google+

  7. #7
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    Yes, that's a bit in your explanation that confused me initially. The "so ..." is general, not related at all to the very next statement you wrote. I was still missing the trivial partial derivative step!

    Thanks,

    i
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