I was doing some practice problems that my professor had sent us and I have not been able to figure out one of them. The given equation is:
$-y^2dx +x^2dy = 0$
He then asks us to verify that:
$ u(x, y) = \frac{1}{(x-y)^2}$
is an integrating factor.
I multiplied through to get:
$\frac{-y^2}{(x-y)^2}dx + \frac{x^2}{(x-y)^2}dy = 0$
However, the partial derivatives of these do not equal each other so I am a bit confused...
On
You must have made a mistake in calculation. Since you do not include your work, we can't point that out. In any case, both partials are $$\frac{-2xy}{(x-y)^3}.$$
It's probably better to first put the differential coefficients into a suitable form. For example, the coefficient of $\mathrm d x$ may be differentiated after put in the form $$-\left(\frac{y}{x-y}\right)^2.$$ Similarly for the second coefficient. It's easier to apply the quotient rule that way without getting into a symbolic mess, which is always a potential circumstance to forget a minus sign somewhere, for example.
$$\frac{\partial}{\partial y}\frac{-y^2}{(x-y)^2}=\frac{\partial}{\partial x}\frac{x^2}{(x-y)^2}=\frac{2xy}{(y-x)^3}$$
Then the partial derivatives is equal.