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Exact Differential EquationEdit

An exact differential equation is a first-order differential equation that can be written in the differential form M(x,y) dx + N(x,y) dy = 0 and recognized, in a region where the functions involved are well-behaved, as the total differential of a scalar function. If there exists a function φ(x,y) whose differential dφ equals M dx + N dy, then the equation is equivalent to dφ = 0, and its general implicit solution is φ(x,y) = C for some constant C. This link to a single potential function makes exact equations a bridge between algebraic manipulation and geometry, since φ serves as a potential whose level curves describe the solution set. The standard criterion for exactness is that the partial derivatives coincide: ∂M/∂y = ∂N/∂x on the domain of interest. When this holds and the domain is simply connected, any solution to M dx + N dy = 0 can be obtained by finding φ such that ∂φ/∂x = M and ∂φ/∂y = N. See also Differential equation and Potential function for related concepts.

The idea of exactness also intersects with physics and multivariable calculus. If M dx + N dy represents a differential form that is exact, the corresponding vector field (M, N) is a gradient field, and line integrals of that field depend only on the endpoints, not on the path taken. This is the path-independence property that characterizes conservative fields and connects to the notion of a potential function φ with M = ∂φ/∂x and N = ∂φ/∂y. See Conservative field and Gradient for related topics, and Line integral for the integral perspective.

Definition and Basic Theory

An exact differential equation is typically written in the form M(x,y) dx + N(x,y) dy = 0, where M and N are functions of two variables with continuous partial derivatives on a region R. The equation is exact on R if there exists a scalar function φ(x,y) such that dφ = M dx + N dy. This is equivalent to the condition ∂M/∂y = ∂N/∂x throughout R. When this holds, the solutions are precisely the level curves of φ, i.e., the curves along which φ(x,y) remains constant.

If an exact equation is given, one can recover the potential function φ by integrating M with respect to x (treating y as a constant) and then determining any y-dependent correction from the remaining information. Concretely, φ(x,y) = ∫ M(x,y) dx + h(y), and then φ_y must equal N, which fixes h'(y). Once φ is known, the general solution is φ(x,y) = C.

Solving Exact Equations

Solving an exact differential equation typically proceeds in two steps: - Show exactness: verify that ∂M/∂y = ∂N/∂x on the region of interest. - Find the potential φ: integrate M with respect to x to obtain φ(x,y) up to a function of y, then determine that function by matching φ_y to N.

Example 1 (exact): Consider M(x,y) = 2xy + x and N(x,y) = x^2. The equation (2xy + x) dx + x^2 dy = 0 is exact because ∂M/∂y = 2x and ∂N/∂x = 2x. Integrating M with respect to x gives φ(x,y) = ∫(2xy + x) dx = x^2 y + x^2/2 + h(y). Differentiating with respect to y yields φ_y = x^2 + h'(y), which must equal N = x^2, so h'(y) = 0 and h(y) is a constant. The potential is φ(x,y) = x^2 y + x^2/2, and the solutions are φ(x,y) = C, i.e., x^2 y + x^2/2 = C.

Example 2 (not exact, but solvable by an integrating factor): Take M(x,y) = y and N(x,y) = 1. Here ∂M/∂y = 1 but ∂N/∂x = 0, so the equation y dx + 1 dy = 0 is not exact. However, there exists an integrating factor μ(x) = e^x because (∂M/∂y − ∂N/∂x)/N = 1/N = 1/1 = 1, a function of x (in this simple case a constant). Multiplying through by μ gives μ M dx + μ N dy = e^x y dx + e^x dy = 0, which is exact. The potential function is φ(x,y) = ∫ e^x y dx = y e^x + h(y); φ_y = e^x + h'(y) must equal μ N = e^x, so h'(y) = 0 and φ = y e^x. The implicit solution is y e^x = C.

Integrating Factors

Not all first-order equations M dx + N dy = 0 are exact as written, but many admit an integrating factor that makes them exact. An integrating factor μ can depend on x alone or on y alone in favorable cases. If μ = μ(x), then the condition for exactness becomes ∂(μ M)/∂y = ∂(μ N)/∂x. This leads to the criterion that (∂M/∂y − ∂N/∂x)/N must be a function of x alone. If μ = μ(y), a similar condition is that (∂N/∂x − ∂M/∂y)/M must be a function of y alone. When such a μ exists, one multiplies the whole equation by μ and then proceeds as in the exact case to find φ and the solution φ = C.

See also Integrating factor, Conservative field, and Gradient for complementary ideas in the broader toolkit of first-order equations and vector fields.

Applications and Connections

Exact differential equations sit at the intersection of analysis and geometry. The equivalence of M dx + N dy to a total differential dφ ties algebraic forms to geometric surfaces: the level curves φ(x,y) = C trace out the integral curves of the equation. This viewpoint clarifies why exact equations encode path-independent information: along any solution curve, the potential φ remains constant.

The concept generalizes beyond two variables and connects to many familiar ideas in multivariable calculus, including line integrals, potential theory, and the study of conservative vector fields. These connections underpin applications ranging from physics (where conservative forces correspond to potential energies) to engineering (where exactness guides the integration of certain conservative models).

See also