Exact Equations

An equation of the form M,dx + N,dy = 0 is exact when \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, because exactness guarantees a potential function F with dF = M,dx + N,dy. Here M = 2xy + 3 and N = x^2 - 1, so \frac{\partial M}{\partial y} = 2x and \frac{\partial N}{\partial x} = 2x, which are equal, confirming the equation is exact. To recover F, integrate M with respect to x: F = \int (2xy + 3),dx = x^2 y + 3x + g(y), where g(y) absorbs any pure-y dependence. Differentiating with respect to y gives \frac{\partial F}{\partial y} = x^2 + g'(y), and matching this to N = x^2 - 1 forces g'(y) = -1, so g(y) = -y. Thus F = x^2 y + 3x - y, and the solution family is x^2 y + 3x - y = C. Option x^2 y - 3x + y = C flips two signs incorrectly. Option xy^2 + 3x - y = C misintegrates the mixed term. Option x^2 y + 3x + y = C keeps the wrong sign for g(y). This is the canonical exact-equation method. As a final check, dF = (2xy+3)dx + (x^2-1)dy reproduces the original equation precisely.