Looking for the name of a class of ODE:

  • Thread starter Thread starter Monochrome
  • Start date Start date
  • Tags Tags
    Class Ode
Click For Summary

Homework Help Overview

The discussion revolves around identifying a specific class of ordinary differential equations (ODEs) characterized by the form M(x,y)y'(x) + N(x,y) = 0, where there exists a function φ(x,y) satisfying certain partial derivative conditions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the classification of the differential equation as an exact equation and discuss the implications of the associated potential function in the context of conservative force fields.

Discussion Status

The conversation includes clarifications regarding typographical errors and the application of the chain rule in the context of partial derivatives. Participants are engaging in a productive exploration of the topic without reaching a definitive conclusion.

Contextual Notes

There is a mention of a typo in the notation, which raises questions about the clarity of the mathematical expressions used. The discussion also highlights the relationship between the differential equation and concepts from physics, specifically conservative force fields.

Monochrome
Messages
11
Reaction score
0

Homework Statement



[tex]M(x,y)y^{'}(x) + N(x,y) = 0[/tex]
There exists:
[tex]\phi(x,y)[/tex]
Such that
[tex]\frac{\partial\phi(x,y)}{\partial x}=N(x,y)[/tex]

[tex]\frac{\partial\phi(x,y)}{\partial y}=M(x,y)[/tex]

I'm not looking for a solution to anything particular to this but I can't find the type in my notes and I can't google it unless I know the name.
 
Physics news on Phys.org
That is an exact equation since
[tex]d\phi= M(x,y)dy+ NIx,y)dx[/itex]<br /> is an exact differential.<br /> <br /> Of course, since the differential equation says [itex]d\phi= 0[/itex], [itex]\phi(x,y)= 0[/itex] is the general solution.[/tex]
 
By the way, in physics, such a differential would correspond to a "conservative force field" and the function [itex]\phi[/itex] would be the "potential function".
 
Thanks.
Just a question: The I after the N is a typo yes? And how did you get [itex]d\phi= 0[/itex]?
 
Last edited:
Yex, that was a typo- my finger was aiming at "("!

Since [itex]\phi[/itex] is a function of both x and y, [itex]d\phi /dx[/itex] would make no sense. By the chain rule, if x and y are functions of some third variable, t,
[tex]\frac{d\phi}{dt}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}[/tex]
or, in differential notation,
[tex]d\phi= \frac{\partial \phi}{\partial x}dx+ \frac{\partial \phi}{\partial y}dy[/tex]
 
HallsofIvy said:
Yex, that was a typo-

:smile:

Thanks I forgot that this was dealing with partials. It makes sense now.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
3K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K