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, \(\displaystyle f(x,y)= 2xy+ y^2/2+ C\).
