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KFC
#1
Aug6-10, 10:32 PM
P: 369
Hi there,
I am reading an introduction on solving ODE, there the exact equation is mentioned. Suppose a ODE is in the following form

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

assume there is a function [tex]F(x, y)[/tex] such that

[tex]\frac{\partial F(x, y)}{\partial x} = M(x, y) + N(x, y)\frac{dy}{dx}[/tex]

Hence,
[tex]\frac{\partial F(x, y)}{\partial x} = 0[/tex]

The text says that the above equation imply that [tex]F(x, y) = \text{Const}[/tex]

But here are my doubts

1) Let's assume the above equation is true, so does it mean [tex]F(x, y)[/tex] has no way to be a function of x if the original equation is exact?

2) What about if [tex]F(x,y) = g(y)[/tex], in this case, we don't know the exact form of y, but we know that y is depending on x, so can we say [tex]F(x,y)=g(y)[/tex] is implicitly depending on x? If it is true, how can we conclude that [tex]F(x, y) = \text{Const}[/tex] instead of some functions of y?

Well, you might find what I am asking is vague. What I actually means is if [tex]F(x,y)=g(y)[/tex], so can we safely say that

[tex]\frac{\partial F(x, y)}{\partial x} = \frac{\partial g(y)}{\partial x} = 0[/tex]

It is quite confusing to use the term 'implicitly'! Because we do know that y=y(x), so how come we put
[tex]\frac{\partial g(y)}{\partial x} = 0[/tex] instead of [tex]\frac{\partial g(y)}{\partial y}\frac{dy}{dx} = 0[/tex] ???
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