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Exact Equation - theory

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.
 

OlderDan

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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
 

James R

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[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]
 
M

MalleusScientiarum

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?
 

HallsofIvy

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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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