Differentials

We have that (y^2 dx + x(y^2) dy) is not an exact differential. The question is: write this differential in the form g df, for functions f(x,y) and g(y) to be determined.

I just don't get this question and haven't come across it before. Any help is much appreciated.

We have that (y^2 dx + x(y^2) dy) is not an exact differential. The question is: write this differential in the form g df, for functions f(x,y) and g(y) to be determined.

I just don't get this question and haven't come across it before. Any help is much appreciated.

so we have (I'll use & instead of curly d) df=(&f/&x)dx+(&f/&y)dy

and g(y) is some function of y

so it seems we should be able to factor out g(y) and obtain an exact differential but that doesn't seem to be possible so are you sure that you typed out the question correctly?

expanding df in the coordinate (dual)basis and equating coefficients gives:
&f/&x=y^2/g(y) and &f/&y=xy^2/g(y)
note that &f/&x is only a function of y

Here is the question (the last part) so you can see it for yourself:





Right... sorry that's not clear! What is & here?

to be clear:
$gdf=g \dfrac{\partial f}{\partial x}dx+g\dfrac{\partial f}{\partial y}dy$
Differentials at a point form a vector space (the cotangent space) so we may equate coefficients with $y^2 dx + x(y^2) dy$ giving:
$g \dfrac{\partial f}{\partial x}=y^2[br]g\dfrac{\partial f}{\partial y}=xy^2$
To deduce a form for f, note that it's partial derivative wrt x only depends on y.

Sorry for this late reply!

So f(x,y) = x h(y), some function h depending only on y.

=> gh = y^2, AND g(dh/dy) = y^2.

Is that right? If so I can't think of a h(y) that satisfies the above...