exact equation
Some first-order equations are secretly already solved — they are the total differential of some hidden function, and we just have to recognize it. Write the equation in the form M(x, y) dx + N(x, y) dy = 0. If the left side is the differential of a single potential function F(x, y), then dF = 0, which means F is constant, and the solution is simply the level curves F(x, y) = C. The equation is called exact when this lucky situation holds.
There is a clean test. The left side equals dF exactly when M is the partial of F with respect to x and N is the partial of F with respect to y; by the equality of mixed partial derivatives, this forces the condition that the partial of M with respect to y equals the partial of N with respect to x. So you check that one equality. If it holds, you reconstruct F by integrating M in x (treating y as constant) and then fixing the y-only piece so that the partial of F with respect to y matches N. The answer is F(x, y) = C — no further integration of the solution needed.
Exact equations are the differential-equations face of conservative vector fields and path-independence from vector calculus: 'exact' is exactly the statement that M dx + N dy is an exact differential, the same condition that makes a line integral path-independent. This is why exactness matters beyond a clever trick — it is the same closed-equals-exact idea that runs through potentials in physics. And when an equation is not exact, the next term, the integrating factor, exists precisely to multiply it into exactness.
(2*x + y) dx + (x + 2*y) dy = 0. Here partial of M w.r.t. y is 1 and partial of N w.r.t. x is 1 — equal, so it is exact. Reconstructing gives F = x^2 + x*y + y^2, and the solution is the family x^2 + x*y + y^2 = C.
Check the mixed-partial test; if it passes, the solution is the level curves F = C.
The exactness test (partial M / partial y = partial N / partial x) being equivalent to 'comes from a potential F' requires the domain to be simply connected (no holes). On a region with a hole the test can pass locally yet no single global F exists — the same subtlety as for conservative fields.