Why are exact ODEs called exact?

**iota** · December 24th 2009, 02:29 AM

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!

**thelostchild** · December 25th 2009, 12:12 AM

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

Merry Christmas!

**HallsofIvy** · December 26th 2009, 02:57 AM

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 $\omega= d\alpha$ .

**iota** · December 27th 2009, 05:15 AM

Originally Posted by HallsofIvy

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

**HallsofIvy** · December 27th 2009, 05:23 AM

Originally Posted by iota

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

**HallsofIvy** · December 27th 2009, 05:23 AM

Originally Posted by iota

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

**iota** · December 27th 2009, 09:48 AM

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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