What is the relationship between exact equations and partial derivatives?

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An exact equation is expressed as M(x,y) + N(x,y) y' = 0, where M and N are partial derivatives of a function ψ(x,y). The relationship between exact equations and partial derivatives is clarified through the total differential, showing that the derivative of ψ with respect to x can be represented as the sum of its partial derivatives. This leads to the conclusion that if y is a function of x, the equation can be simplified to dψ/dx = 0. The discussion emphasizes that using partial derivatives helps in expressing the sum of differentials as a single differential, simplifying the problem. Understanding this relationship is crucial for solving differential equations effectively.
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An exact equation has the form

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

where

M(x,y) = \frac{\partial \psi}{\partial x} (x,y)

and

N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}

If y=\phi (x) and \psi (x,y) = c, then

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{.}

I can't understand this:

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

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

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

df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy

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

If you divide by dx you have

\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}

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

Well, \psi is a function of x and y, so

\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{\partial y}{\partial x}

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

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

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

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

\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \mbox{.}
 
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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