There is a remote possibility that this is not an intentional exercise, as my book actually states the equation as 2y dx + 3x dx = 0. I have assumed this to be a misprint. This exercise appears in the section on exact differential equations.Consider the differential equation 2y dx + 3x dy = 0. Determine conditions on a and b so that u(x, y) = (x^a)(y^b) is an integrating factor. Find a particular integrating factor and use it to solve the differential equation.

The scalar curl of the equation is ∂(3x)/∂x - ∂(2y)/∂y = 1. Multiplying by u(x, y) gives 2(x^a)(y^(b + 1)) dx + 3(x^(a + 1))(y^b) dy = 0, which has a scalar curl of (3a - 2b + 5)(x^a)(y^b). (3a - 2b + 5)(x^a)(y^b) = 0 could maybe sorta qualify as conditions on a and b that yield an integrating factor, but only in a very implicit and not very insightful way, such that I don't think this is the answer to the first part of the question.

The equation is homogeneous, so it might be appropriate to make the substitution y = xv; dy = dx v + x dv. This transforms the equation into 5xv dx + 3x^2 dv = 0, which has a scalar curl of x. Dividing through by x to get 5v dx + 3x dv = 0 yields a scalar curl of -2. These two substituted equations multiplied by u(x, y) have scalar curls of (3a - 2b + 1)(x^(a + b + 1))(v^b) and (3a - 2b - 2)(x^(a + b))(v^b), respectively. The integrating factor u(x, y) doesn't seem to simplify things in any case.

Obviously (x^-1)(y^-1) is a particular integrating factor, as it separates the variables, but I'm not sure how to arrive at this conclusion without that insight, given a general integrating factor, as implied by the problem.