integrating factor that restores exactness
/ mu(x), Greek 'mew' /
Suppose the exactness test fails — partial M / partial y is not equal to partial N / partial x — so the equation has no potential as written. All is not lost: you may be able to multiply the whole equation by a cleverly chosen function mu(x, y) so that the NEW coefficients pass the test. That magic multiplier is an integrating factor, written mu, and it converts a non-exact equation into an exact one without changing its solution curves.
After multiplying, the equation is (mu M) dx + (mu N) dy = 0, and you want partial(mu M)/partial y = partial(mu N)/partial x. In general this is a hard partial differential equation for mu, but two friendly special cases save the day. If the combination (partial M/partial y - partial N/partial x)/N depends on x alone, then mu is a function of x alone, found by mu(x) = e^(integral of that expression dx). Symmetrically, if (partial N/partial x - partial M/partial y)/M depends on y alone, then mu is a function of y alone. In either case you integrate one ordinary function, multiply through by mu, and now recover F as usual.
This is the same idea as the integrating factor for a first-order linear equation, generalized: there the factor e^(integral of p dx) collapses the left side into the derivative of a product; here mu collapses M dx + N dy into a genuine total differential dF. The honest limit is that there is no universal recipe — when neither special case applies, finding mu can be as hard as solving the original equation, which is why the elementary toolkit only reaches so far.
The equation y dx + (2x - y e^y) dy = 0 is not exact (partial M/partial y = 1, partial N/partial x = 2). Test the y-only case: (partial N/partial x - partial M/partial y)/M = (2 - 1)/y = 1/y, which depends on y alone, so mu(y) = e^(integral of 1/y dy) = y. Multiplying gives y^2 dx + (2xy - y^2 e^y) dy = 0, which is now exact.
When the mismatch simplifies to a one-variable expression, mu is just e to its integral.
An integrating factor can introduce or remove solutions where mu = 0 or where mu blows up. Multiplying by mu is only an equivalence on the region where mu is finite and nonzero, so check the curves you may have added or lost there.