Exact diff. eqns.

[patter2809]

If M = ∂f/∂x and N = ∂f/∂y, we are told Mdx + Ndy = 0 if and only if ∂M/∂y = ∂N/∂x.

1) Mdx+Ndy = ∂f/∂x . dx + ∂f/∂y . dy = 0

∂f/∂x = -∂f/∂y . dy/dx

∂f/∂x = ∂f/∂y x dy/dx because y = f(x), but that gives Ndy/dx = -N dy/dx.

Is ∂f/∂x = ∂f/∂y x dy/dx not valid or have I gone wrong somewhere else?
Thanks in advance.

 

[HallsofIvy]

If M = ∂f/∂x and N = ∂f/∂y, we are told Mdx + Ndy = 0 if and only if ∂M/∂y = ∂N/∂x.

I wouldn't put it that way. If $\displaystyle M= \partial f/\partial x$ and $\displaystyle N= \partial f/\partial y$, it follows immediately that \(\displaystyle \partial M/\partial y= \partial N/\partial x. \(\displaystyle Mdx+ Ndy= 0\) is true if and only if f(x,y)= Constant for all x and y.

What you are probably thinking of is the other way around. Given two functions M and N, then Mdx+ Ndy= df if and only if $\displaystyle M= \partial f/\partial x$ and $\displaystyle N= \partial N/\partial y$.

1) Mdx+Ndy = ∂f/∂x . dx + ∂f/∂y . dy = 0

∂f/∂x = -∂f/∂y . dy/dx

dy/dx has no meaning unless y is a function of x. Do you mean this on a specific curve f(x, y)= costant?

∂f/∂x = ∂f/∂y x dy/dx because y = f(x), but that gives Ndy/dx = -N dy/dx.

Is ∂f/∂x = ∂f/∂y x dy/dx not valid or have I gone wrong somewhere else?
Thanks in advance.

You have suddenly started assuming that y= f(x). Where did you get that from?\)

 

[patter2809]

Yes, sorry that's what i meant. I couldn't get to Mdx + Ndy = 0 based on those definitions of M, and N from f(x,y) = c.

To clarify, y = h(x) so dy/dx would be defined as you said.

 

[HallsofIvy]

If M = ∂f/∂x and N = ∂f/∂y, we are told Mdx + Ndy = 0 if and only if ∂M/∂y = ∂N/∂x.

No, you are NOT told that. What you are incorrectly remembering is that if $\displaystyle M= \frac{\partial f}{\partial x}$ and $\displaystyle N= \frac{\partial f}{\partial x}$ and the second derivatives are continuous then $\displaystyle \frac{\partial M}{\partial x}= \frac{\partial f}{\partial y}$. That's true because of the "equality of mixed derivatives" in this case: $\displaystyle \frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}$ where the changed orders of "x" and "y" indicate the different orders of differentiation. That has nothing to do with Mdx+ Ndy= 0

For any differentiable function, f, of variables x and y, the chain rule asserts that $\displaystyle \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$ or, in "differential" term, $\displaystyle df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy$. Saying that differential is 0 means that f is a constant.

1) Mdx+Ndy = ∂f/∂x . dx + ∂f/∂y . dy = 0

∂f/∂x = -∂f/∂y . dy/dx

∂f/∂x = ∂f/∂y x dy/dx because y = f(x), but that gives Ndy/dx = -N dy/dx.

Is ∂f/∂x = ∂f/∂y x dy/dx not valid or have I gone wrong somewhere else?
Thanks in advance.

Your basic understanding is wrong. $\displaystyle M_y= N_x$ says that there exist f(x,y) such that $\displaystyle f_x= N$ and $\displaystyle f_y= M$. That together with Mdx+ Mdy= 0 says that f(x,y)= constant.

For example, if 2ydx+ (2x+ y)dy= 0 (note that (2y)_y= 2= (2x+ y)_x) then $\displaystyle \frac{\partial f}{\partial x}= 2y$ so that f(x, y)= 2xy+ g(y). Because the "partial derivative with respect to x" treats y as a constant, the "constant of integration" may be a function of y. Now, differentiating that with respect to y, we have $\displaystyle \frac{\partial f}{\partial y}= 2x+ g'$ and we know that must be equal to 2x+ y. The "2x" terms cancel (that is because of the $\displaystyle M_y= N_x$) so that we must have $\displaystyle g'= y$ and so $\displaystyle g(y)= y^2/2+ C$ where, since g is a function of y only, C really is a constant. Putting that into the equation, $\displaystyle f(x,y)= 2xy+ y^2/2+ C$.