What is the relationship between exact equations and partial derivatives?

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The discussion revolves around the relationship between exact equations and partial derivatives, specifically focusing on the form of exact equations and their implications in calculus. Participants are exploring the definitions and properties of exact differentials and their connection to total differentials.

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  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the expression relating partial derivatives to the total derivative of a function. There are questions about the definitions of exact differentials and how they relate to the conversion of ordinary differential equations.

Discussion Status

Some participants have provided insights into the definitions of exact differentials and the reasoning behind the expressions used. However, there is still uncertainty regarding the practical implications of converting between types of equations and the overall understanding of the topic.

Contextual Notes

There is mention of potential confusion regarding the distinction between ordinary and partial differential equations, as well as the assumptions underlying the definitions being discussed.

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An exact equation has the form

[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

where

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

and

[tex]N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}[/tex]

If [tex]y=\phi (x)[/tex] and [tex]\psi (x,y) = c[/tex], then

[tex]M(x,y) + N(x,y) \: y^{\prime} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right] = 0 \mbox{.}[/tex]

I can't understand this:

[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]

Any help is highly appreciated.
 
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thiago_j said:
I can't understand this:

[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]
From the definition of an "exact differential" or "total differential" for a function f(x,y) whose integral is path independent

[tex]df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy[/tex]

http://mathworld.wolfram.com/ExactDifferential.html

If you divide by dx you have

[tex]\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}[/tex]

and since you have y as a function of x, there is a derivative of y wrt x
 
[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]

Well, [itex]\psi[/itex] is a function of x and y, so

[tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{\partial y}{\partial x}[/tex]

Now put [itex]y=\phi (x)[/itex], and that's all there is to it.
 
Thank you, guys! Thus, we have

[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

[tex]d\psi = M(x,y) \: dx + N(x,y) \: dy = 0[/tex]

[tex]\frac{d\psi}{dx} = M(x,y) \: + N(x,y) \: \frac{dy}{dx} = 0[/tex]

[tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \mbox{.}[/tex]
 
I'm not exactly sure of how pragmatic it is to convert an ordinary differential equation into a partial differential equation. Maybe I miss something?
 
No one changed an ordinary differential equation into a partial differential equation. They just used partial derivatives to show show that the sum of two differentials can, in some circumstances, be written as a single differential- which makes the problem very easy.
 

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