# What is an Exact DE?

member 392791
I am having confusion, so from what I understand for an exact DE

dy/dx = some DE and you can manipulate it so that (y/x) appears and substitute that for v when you say y=vx, then just use separation of variables and solve.

However, what is the significance of an exact DE, why is it useful, or is it just a way that someone has figured out how to solve and bares no meaning beyond being a method to solve that type of DE?

Simon Bridge
Homework Helper
It's a classification of a DE according to it's properties - the label is applied without reference to the method of solving it. i.e. you don't need to see if it can be solved by the method for exact DEs to discover it is exact.

It's like polynomials of order 2 are called "quadratics" - quadratics have particular characteristics and you have a standard set of tools for dealing with them.

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The nomenclature of "exact differential equation" refers to the exact derivative of a function.
http://en.wikipedia.org/wiki/Exact_differential_equation

HallsofIvy
Homework Helper
Back in Calculus, you should have seen the concept of an "exact differential". If we have a function, f(x, y), with x and y both depending on the parameter, t, we could replace them with their expression in terms of t and find df/dt. Or, we could use the "chain rule":
$$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$$
We can then change to "differential form"
$$df= \frac{df}{dt}dt= (\frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt})dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy$$
Notice that the last form,
$$df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy$$
has no mention of "t" and is true irrespective of any parameter.

Of course, if I have an equation that says "df= 0" then I can immediately write "f(x,y)= C" for some constant C.

But I can write g(x,y)dx+ h(x,y)dy that are NOT, in fact, "exact differentials".
If $df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy= g(x,y)dx+ h(x, y)dy$, and g and h have, themselves, continuous derivatives, we must have $\partial g/\partial y= \partial^2 f/\partial x\partial y= \partial h/\partial x$
If that is NOT true, gdx+ hdy is just "pretending" to be a differential- it not an "exact differential". If it is true, then gdx+ hdy= 0 is an "exact differential equation" and, once we find the correct f(x,y), we can write the general solution to the equation as f(x,y)= Constant.

member 392791
I feel dumb haha, I said what is an exact DE but what I described is homogeneous. I guess what is both really

HallsofIvy
Homework Helper
The phrase "homogeneous differential equation" is used in two distinctly different ways. A first order differential equation can be written in the form dy/dx= f(x, y) for some function f. The function is said to be "homogeneous" of degree n if $f(\lambda x, \lambda y)= \lambda^n f(x, y)$. The differential equation dy/dx= f(x,y) is "homogeneous" if and only if f is "homogeneous".

But "homogeneous" has a different meaning for higher order linear equations. A linear differential equation can be written in the form $f_n(x)d^ny/dx^n+ f_{n-1}(x)d^{n-1}y/dx^{n-1}+ \cdot\cdot\cdot+ f_1(x)dy/dx+ f_0(x)y= g(x)$. Such an equation is "homogenous[/b] if and only if the function on the right (the only one not multiply the function y or a derivative of it) is 0.

Those definitions can be found in any differential equations textbook or on Wikipedia at http://en.wikipedia.org/wiki/Homogeneous_differential_equation.

member 392791
I get these definitions (I've read the wikipedia articles), but I don't feel like I really understand them on anymore than a superficial level. Yeah I can spout off the definition and carry out a calculation, but do I really know the heart and soul of it??

For the exact DE, what is the potential function supposed to represent, maybe in a real life situation (mathematics rigorous definitions are incomprehensible to me).

Simon Bridge