Exact Equations

Suppose a curve in the plane is defined implicitly by an equation like x2y+y3=Cx^2 y + y^3 = C. There is one level curve for each constant CC. Differentiating implicitly produces a first-order differential equation. The natural question is this: given only that differential equation, how would you reconstruct the hidden function F(x,y)=x2y+y3F(x,y)=x^2y+y^3? This function's level sets are what the equation describes. This is the reverse problem solved by exact equations. It matters because a huge class of first-order differential equations that are not separable can still be solved in closed form. This works once they are recognized as disguised total differentials. The big picture is that a differential equation M dx+N dy=0M\,dx+N\,dy=0 is really a statement about the differential of some scalar function F(x,y)F(x,y). Solving the equation amounts to finding FF so that the solutions are simply the level curves F(x,y)=CF(x,y)=C. This turns a differential equation problem into a problem about partial derivatives and antiderivatives. These are tools already familiar from multivariable calculus. This article gives the precise definition of exactness. It proves the exactness test My=NxM_y=N_x rigorously, in both directions, with correct hypotheses. It works through the reconstruction algorithm on two genuinely asymmetric examples. It also shows exactly what fails when the domain is not simply connected. This sets up the natural next tool, the Integrating Factor, for equations that fail the test as written.

Key takeaways
  • An equation M dx+N dy=0M\,dx + N\,dy = 0 is exact on a region if M dx+N dyM\,dx+N\,dy is the total differential dFdF of some function FF.
  • On a simply connected region with M,N∈C1M,N \in C^1, exactness is equivalent to the mixed-partial condition My=NxM_y = N_x.
  • Solutions are the implicit level curves F(x,y)=CF(x,y)=C, so no separation of variables is needed once FF is found.
  • Recovering FF is a two-step antiderivative process: integrate MM in xx, then match ∂F/∂y\partial F/\partial y against NN.
  • The condition My=NxM_y=N_x can fail to imply exactness if the domain has a "hole" — simple connectedness is essential.
  • Key idea: exactness converts a differential equation into a calculus problem — reconstructing a potential function from its two partial derivatives.

Where this lives today

Exact equations are the differential-equations shadow of conservative vector fields in physics: a force field F=(M,N)\mathbf{F}=(M,N) is conservative exactly when M dx+N dyM\,dx+N\,dy is an exact differential, and the potential function FF recovered here is (up to sign) the potential energy function from mechanics and electromagnetism. In thermodynamics, exact differentials distinguish state functions (internal energy, entropy) from path-dependent quantities (heat, work), and the equality My=NxM_y=N_x reappears as a Maxwell relation. In circuit theory and control systems, exactness underlies conserved quantities and Lyapunov-function constructions for planar systems. Computationally, computer algebra systems that solve ODEs symbolically run the My=NxM_y=N_x test as a first triage step before attempting more elaborate methods, and numerical schemes for Hamiltonian systems are built to preserve the exact/conservative structure explicitly. The same reconstruction algorithm — integrate one partial, differentiate to match the other — is the template later reused, with a corrective multiplier, when the equation fails the test as given, leading directly into the method of Integrating Factors.

Ask the tutor: How do I find an integrating factor when My≠NxM_y \neq N_x, and how do I know whether it depends only on xx, only on yy, or on both?

DefinitionExact Differential Equation

Let D⊆R2D \subseteq \mathbb{R}^2 be an open region, and let M,N:D→RM, N : D \to \mathbb{R} be continuous functions. The first-order differential equation

M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0

is called exact on DD if there exists a function F:D→RF : D \to \mathbb{R}, differentiable on DD, such that

∂F∂x(x,y)=M(x,y),∂F∂y(x,y)=N(x,y)for all (x,y)∈D.\frac{\partial F}{\partial x}(x,y) = M(x,y), \qquad \frac{\partial F}{\partial y}(x,y) = N(x,y) \quad \text{for all } (x,y) \in D.

Such an FF is called a potential function for the equation. If y=ϕ(x)y=\phi(x) is a differentiable function satisfying the equation implicitly, the chain rule gives

ddxF(x,ϕ(x))=Fx+Fyϕ′(x)=M+Nϕ′=0,\frac{d}{dx}F(x,\phi(x)) = F_x + F_y \phi'(x) = M + N\phi' = 0,

so F(x,ϕ(x))≡CF(x,\phi(x)) \equiv C for some constant CC. Consequently, every solution curve lies on a level set F(x,y)=CF(x,y)=C, and the general solution of an exact equation is given implicitly by

F(x,y)=C,C∈R.F(x,y) = C, \qquad C \in \mathbb{R}.

Intuition

Think of F(x,y)F(x,y) as a height function over the plane, like elevation on a topographic map. Its level curves F(x,y)=CF(x,y)=C are the contour lines. The gradient ∇F=(Fx,Fy)=(M,N)\nabla F = (F_x, F_y) = (M,N) is a vector field that, at every point, points in the direction of steepest ascent — and it is always perpendicular to the contour line through that point. A differential equation M dx+N dy=0M\,dx + N\,dy=0 is exactly the statement "move in a direction (dx,dy)(dx,dy) tangent to the contour," because dF=M dx+N dy=0dF = M\,dx+N\,dy=0 says the height does not change to first order along that displacement. So solving an exact equation is literally the act of walking along contour lines of a hidden landscape FF; the whole problem is to reconstruct the landscape from the vector field (M,N)=∇F(M,N)=\nabla F that describes its slopes. The condition My=NxM_y = N_x is the familiar equality of mixed partial derivatives, Fxy=FyxF_{xy} = F_{yx}, viewed from the outside: it is a necessary signature that (M,N)(M,N) really is a gradient field, before we've found FF. Geometrically, My=NxM_y=N_x says the vector field (M,N)(M,N) has zero "rotation" (its scalar curl vanishes) — a gradient field can never rotate around a point, since going all the way around any small loop back to the start cannot change the height. The theorem below makes this precise and shows the converse holds provided the domain has no holes to trap a nonzero circulation.

Figure
Level curves of F(x,y)=x2y+y3F(x,y)=x^2y+y^3 and the gradient field (M,N)=(2xy, x2+3y2)(M,N)=(2xy,\,x^2+3y^2)
x y F = -6 F = -3 F = 0 ∇F = (M,N) contours ⟂ gradient
TheoremExactness Criterion

Let D⊆R2D \subseteq \mathbb{R}^2 be an open, simply connected region, and let M(x,y)M(x,y) and N(x,y)N(x,y) be functions with continuous partial derivatives MyM_y and NxN_x on DD. Then the differential equation

M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0

is exact on DD if and only if

∂M∂y(x,y)=∂N∂x(x,y)for all (x,y)∈D.\frac{\partial M}{\partial y}(x,y) = \frac{\partial N}{\partial x}(x,y) \qquad \text{for all } (x,y)\in D.

Moreover, when this condition holds, a potential function FF can be constructed explicitly, and the general solution of the equation is F(x,y)=CF(x,y)=C.

Proof

We prove both directions. The proof technique is direct construction (sufficiency) and a computation using equality of mixed partials (necessity).

(⇒\Rightarrow) Necessity: exactness implies My=NxM_y=N_x. Suppose the equation is exact on DD, so there is FF with Fx=MF_x = M and Fy=NF_y=N throughout DD. Since MyM_y is assumed continuous, Fx=MF_x = M has a continuous yy-partial, i.e. Fxy=MyF_{xy}=M_y is continuous on DD. Likewise Fyx=NxF_{yx}=N_x is continuous on DD. By Clairaut's theorem (equality of mixed partial derivatives), if FxyF_{xy} and FyxF_{yx} both exist and are continuous on an open set, then

Fxy=Fyxon that set.F_{xy} = F_{yx} \quad \text{on that set}.

Applying this with Fxy=MyF_{xy}=M_y and Fyx=NxF_{yx}=N_x gives My=NxM_y = N_x on DD, as required. This direction uses only that DD is open (Clairaut's theorem needs no topological hypothesis on DD beyond openness, and does not require simple connectedness).

(⇐\Leftarrow) Sufficiency: My=NxM_y=N_x implies exactness, via explicit construction. Assume now My=NxM_y = N_x on DD and DD is open and simply connected. Fix a base point (x0,y0)∈D(x_0,y_0)\in D. We construct FF directly by partial integration and verify it works; this is the standard algorithm, and simple connectedness is what guarantees the construction is consistent (single-valued) throughout DD. Step 1 (partial antiderivative in xx). Define, for (x,y)(x,y) in a neighborhood where a straight path from (x0,y)(x_0,y) to (x,y)(x,y) stays in DD,

F(x,y)=∫x0xM(s,y) ds+g(y)F(x,y) = \int_{x_0}^{x} M(s,y)\,ds + g(y)

for an as-yet-undetermined function g(y)g(y). By the Fundamental Theorem of Calculus (differentiating under the integral sign, valid since MM is continuous),

Fx(x,y)=M(x,y),F_x(x,y) = M(x,y),

which matches the required condition automatically, for any choice of gg. Step 2 (determine gg by matching Fy=NF_y=N). Differentiate FF with respect to yy, using that MyM_y is continuous so differentiation under the integral sign is valid (Leibniz's rule):

Fy(x,y)=∫x0xMy(s,y) ds+g′(y).F_y(x,y) = \int_{x_0}^{x} M_y(s,y)\,ds + g'(y).

Using the hypothesis My=NxM_y = N_x pointwise inside the integral,

Fy(x,y)=∫x0xNx(s,y) ds+g′(y)=N(x,y)−N(x0,y)+g′(y),F_y(x,y) = \int_{x_0}^{x} N_x(s,y)\,ds + g'(y) = N(x,y) - N(x_0,y) + g'(y),

by the Fundamental Theorem of Calculus applied to Nx(⋅,y)N_x(\cdot,y) over [x0,x][x_0,x]. For FyF_y to equal N(x,y)N(x,y) exactly, we require

g′(y)=N(x0,y).g'(y) = N(x_0,y).

This is an equation in yy alone (the right side does not depend on xx, which is precisely what the hypothesis My=NxM_y=N_x guaranteed), so it can be solved by direct antidifferentiation:

g(y)=∫y0yN(x0,t) dt.g(y) = \int_{y_0}^{y} N(x_0,t)\,dt.

Step 3 (assemble and verify). Define

F(x,y)=∫x0xM(s,y) ds+∫y0yN(x0,t) dt.F(x,y) = \int_{x_0}^{x} M(s,y)\,ds + \int_{y_0}^{y} N(x_0,t)\,dt.

By Steps 1 and 2, Fx=MF_x = M and Fy=NF_y = N identically on the region where this construction is valid. This confirms FF is a genuine potential function there, so the equation is exact there. Step 4 (globalizing via simple connectedness). The construction above is local: it produces FF correctly on any rectangle or star-shaped neighborhood of (x0,y0)(x_0,y_0) contained in DD. To extend FF consistently over all of DD, note that the vector field (M,N)(M,N) satisfying My=NxM_y=N_x is exactly a vector field with zero scalar curl. A standard fact from vector calculus (a consequence of Green's theorem) states: if DD is simply connected and (M,N)∈C1(D)(M,N)\in C^1(D) with My=NxM_y=N_x, then the line integral

∫γM dx+N dy\int_\gamma M\,dx + N\,dy

is path-independent in DD — any two paths in DD with the same endpoints give the same integral, because the difference of the integrals over two such paths equals the integral over a closed loop enclosing a simply connected sub-region, which by Green's theorem equals

∮(M dx+N dy)=∬enclosed(Nx−My) dA=0.\oint (M\,dx+N\,dy) = \iint_{\text{enclosed}} (N_x - M_y)\,dA = 0.

(Simple connectedness of DD is exactly what guarantees the region enclosed by any closed loop in DD is itself contained in DD, so Green's theorem applies with the correct hypotheses.) Path-independence lets us define, unambiguously for every (x,y)∈D(x,y) \in D,

F(x,y)=∫(x0,y0)(x,y)M dx+N dy,F(x,y) = \int_{(x_0,y_0)}^{(x,y)} M\,dx+N\,dy,

the integral taken along any path in DD from (x0,y0)(x_0,y_0) to (x,y)(x,y). Choosing the specific path that goes horizontally from (x0,y0)(x_0,y_0) to (x0,y)(x_0,y) and then vertically to (x,y)(x,y) reproduces exactly the formula from Step 3 (up to relabeling), and the same differentiation argument shows Fx=MF_x=M, Fy=NF_y=N on all of DD. Hence FF is a well-defined potential function on the entire region DD, proving exactness there.

Conclusion. Both directions are established: on an open, simply connected DD with M,N∈C1(D)M,N\in C^1(D), exactness of M dx+N dy=0M\,dx+N\,dy=0 holds if and only if My=NxM_y=N_x throughout DD, and in that case FF constructed above satisfies F(x,y)=CF(x,y)=C as the general implicit solution, by the remark following the definition. ■\blacksquare

∎

Worked examples

Example 1 (asymmetric coefficients, requiring genuine bookkeeping). Solve

(3x2y+2xy2−y) dx+(x3+2x2y−x+2y) dy=0.(3x^2y + 2xy^2 - y)\,dx + (x^3 + 2x^2y - x + 2y)\,dy = 0.

Step 1: Identify MM and NN and test exactness. Here

M(x,y)=3x2y+2xy2−y,N(x,y)=x3+2x2y−x+2y.M(x,y) = 3x^2y+2xy^2-y, \qquad N(x,y) = x^3+2x^2y-x+2y.

Compute the cross partials:

My=3x2+4xy−1,Nx=3x2+4xy−1.M_y = 3x^2 + 4xy - 1, \qquad N_x = 3x^2+4xy-1.

Since My=NxM_y = N_x on all of R2\mathbb{R}^2 (an open, simply connected region, and both M,NM,N are polynomials, hence C1C^1 everywhere), Exactness Criterion applies and guarantees an exact potential FF. Step 2: Integrate MM in xx, treating yy as constant.

F(x,y)=∫(3x2y+2xy2−y) dx=x3y+x2y2−xy+h(y),F(x,y) = \int (3x^2y+2xy^2-y)\,dx = x^3y + x^2y^2 - xy + h(y),

where h(y)h(y) is an unknown function of yy alone (the "constant" of integration with respect to xx). Step 3: Differentiate with respect to yy and match against NN.

Fy(x,y)=x3+2x2y−x+h′(y).F_y(x,y) = x^3 + 2x^2y - x + h'(y).

Set this equal to N=x3+2x2y−x+2yN = x^3+2x^2y-x+2y:

x3+2x2y−x+h′(y)=x3+2x2y−x+2y⟹h′(y)=2y.x^3+2x^2y-x+h'(y) = x^3+2x^2y-x+2y \quad \Longrightarrow \quad h'(y) = 2y.

Note that all the xx-dependent terms canceled exactly — this cancellation is not a coincidence, it is the content of My=NxM_y=N_x verified in Step 1, and it is exactly what the general proof showed must happen. Step 4: Integrate to find h(y)h(y), and assemble FF.

h(y)=y2+C0,h(y) = y^2 + C_0,

and absorbing the constant into the final answer,

F(x,y)=x3y+x2y2−xy+y2.F(x,y) = x^3y + x^2y^2 - xy + y^2.

Step 5: State the general solution.

x3y+x2y2−xy+y2=C.x^3 y + x^2 y^2 - xy + y^2 = C.

Check. Differentiate implicitly with respect to xx (treating y=y(x)y=y(x)):

3x2y+x3y′+2xy2+2x2yy′−y−xy′+2yy′=0.3x^2y + x^3y' + 2xy^2 + 2x^2yy' - y - xy' + 2yy' = 0.

Grouping y′y' terms:

(3x2y+2xy2−y)+(x3+2x2y−x+2y)y′=0,(3x^2y+2xy^2-y) + (x^3+2x^2y-x+2y)y' = 0,

which reproduces exactly M+Ny′=0M + Ny' = 0, confirming the solution.

Example 2 (an initial value problem, and where the recipe subtly differs). Solve the initial value problem

(exsin⁡y−2ysin⁡x) dx+(excos⁡y+2cos⁡x) dy=0,y(0)=π.(e^{x}\sin y - 2y\sin x)\,dx + (e^{x}\cos y + 2\cos x)\,dy = 0, \qquad y(0) = \pi.

Step 1: Test exactness.

M=exsin⁡y−2ysin⁡x,N=excos⁡y+2cos⁡x.M = e^x\sin y - 2y\sin x, \qquad N = e^x \cos y + 2\cos x.
My=excos⁡y−2sin⁡x,Nx=excos⁡y−2sin⁡x.M_y = e^x \cos y - 2\sin x, \qquad N_x = e^x\cos y - 2\sin x.

Indeed My=NxM_y = N_x everywhere on R2\mathbb{R}^2, which is open and simply connected, and M,N∈C1(R2)M, N \in C^1(\mathbb{R}^2), so Exactness Criterion applies. Step 2: Integrate MM with respect to xx. This time integrating MM directly is efficient:

F(x,y)=∫(exsin⁡y−2ysin⁡x) dx=exsin⁡y+2ycos⁡x+h(y).F(x,y) = \int (e^x \sin y - 2y\sin x)\,dx = e^x\sin y + 2y\cos x + h(y).

(Here ∫−2ysin⁡x dx=2ycos⁡x\int -2y\sin x\,dx = 2y\cos x since yy is held fixed during this integration.) Step 3: Differentiate in yy and match against NN.

Fy=excos⁡y+2cos⁡x+h′(y).F_y = e^x\cos y + 2\cos x + h'(y).

Comparing to N=excos⁡y+2cos⁡xN = e^x\cos y + 2\cos x gives h′(y)=0h'(y) = 0, so h(y)h(y) is a genuine constant — in this example the entire yy-dependence was already captured in Step 2, a useful check that the arithmetic is consistent. Step 4: Assemble FF and apply the initial condition.

F(x,y)=exsin⁡y+2ycos⁡x=C.F(x,y) = e^x \sin y + 2y \cos x = C.

Impose x=0, y=πx=0,\ y=\pi:

e0sin⁡π+2πcos⁡0=0+2π=2π.e^0 \sin \pi + 2\pi \cos 0 = 0 + 2\pi = 2\pi.

So C=2πC = 2\pi, and the solution of the IVP is given implicitly by

exsin⁡y+2ycos⁡x=2π.e^x \sin y + 2y\cos x = 2\pi.

Remark. Unlike a separable equation, this solution cannot be split into "f(y) dy=g(x) dxf(y)\,dy = g(x)\,dx" form — the terms exsin⁡ye^x\sin y and 2ycos⁡x2y\cos x genuinely mix xx and yy. Exactness succeeds precisely because it does not require separability; it only requires recognizing (M,N)(M,N) as a gradient field.

Counterexamplewhy simple connectedness cannot be dropped

Consider the vector field

M(x,y)=−yx2+y2,N(x,y)=xx2+y2,M(x,y) = \frac{-y}{x^2+y^2}, \qquad N(x,y) = \frac{x}{x^2+y^2},

defined and C1C^1 on D=R2∖{(0,0)}D = \mathbb{R}^2 \setminus \{(0,0)\}, the punctured plane. This DD is open but not simply connected — a loop around the origin cannot be contracted to a point without leaving DD. The cross-partial condition holds. A direct computation (quotient rule) gives

My=−(x2+y2)+y(2y)(x2+y2)2=y2−x2(x2+y2)2,Nx=(x2+y2)−x(2x)(x2+y2)2=y2−x2(x2+y2)2.M_y = \frac{-(x^2+y^2) + y(2y)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, \qquad N_x = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.

So My=NxM_y = N_x everywhere on DD. But the equation is not exact on DD. If it were exact, there would be a potential FF with Fx=MF_x=M, Fy=NF_y=N on all of DD, and then by the Fundamental Theorem of Calculus for line integrals, the integral of M dx+N dyM\,dx+N\,dy around any closed loop in DD would be 00 (it would equal F(endpoint)−F(start point)=0F(\text{endpoint}) - F(\text{start point}) = 0 for a closed loop). Test this on the unit circle γ(t)=(cos⁡t,sin⁡t)\gamma(t) = (\cos t, \sin t), t∈[0,2π]t\in[0,2\pi]:

∮γM dx+N dy=∫02π[(−sin⁡t)(−sin⁡t)+(cos⁡t)(cos⁡t)] dt=∫02π1 dt=2π≠0.\oint_\gamma M\,dx+N\,dy = \int_0^{2\pi} \big[(-\sin t)(-\sin t) + (\cos t)(\cos t)\big]\,dt = \int_0^{2\pi} 1\,dt = 2\pi \neq 0.

So no global potential function exists on DD: the equation satisfies My=NxM_y=N_x but is not exact on the punctured plane. (In fact, F=arctan⁡(y/x)F=\arctan(y/x), i.e. the angle θ\theta, works locally away from any ray through the origin, but θ\theta cannot be extended continuously to a single-valued function on all of DD — exactly the obstruction simple connectedness rules out.) This shows the hypothesis "DD simply connected" in Exactness Criterion cannot be dropped: without it, My=NxM_y=N_x is necessary but no longer sufficient for exactness.

Scope, edge cases & common traps

SituationWhat to do
My=NxM_y = N_x on a rectangle or the whole planeApply the theorem directly; such domains are automatically simply connected.
My=NxM_y = N_x but domain has a hole (e.g. origin excluded)Do not assume exactness; check path-independence directly, as in the counterexample.
My≠NxM_y \neq N_xThe equation is not exact as written; look for an Integrating Factor to multiply through and restore exactness.
Constant of integration h(y)h(y) turns out to depend on xxSignals an arithmetic error, since the theorem guarantees the xx-dependence must cancel whenever My=NxM_y=N_x truly holds.
Confusing "exact" with "separable"These are different tests. A separable equation need not be exact and vice versa; exactness is checked via partials, not by attempting a variable split.

The mathematician behind it

Origins · 1 minAlexis Clairaut and how this was discovered
AC
Alexis Clairaut
1713–1765 · 18th-century French mathematician and astronomer

Clairaut proved the equality of mixed partial derivatives underlying the necessity direction of the exactness test, and studied differential equations arising from celestial mechanics, including the singular equations that today bear his name.

History
- 1734 — Clairaut publishes work on differential equations, including what is now called Clairaut's equation.
- 1740 — Clairaut's theorem on the equality of mixed second partial derivatives is established in the form used throughout analysis.
- 18th century — Euler and the Bernoullis develop the calculus of exact differentials in tandem with early work on line integrals and potential functions.
- 19th century — Green's and Cauchy's work on line integrals and the theorem connecting circulation to enclosed area (Green's theorem) makes the topological role of simple connectedness precise.
- Present — the exactness test is standard early material in every differential equations course, immediately motivating the search for integrating factors.
How is exactness different from separability?

Separability asks whether M dx+N dy=0M\,dx+N\,dy=0 can be algebraically rearranged into f(y) dy=g(x) dxf(y)\,dy = g(x)\,dx, so that both sides can be integrated independently, as in Separable Equations. Exactness instead asks whether the pair (M,N)(M,N) is literally the gradient of some scalar function FF, tested via My=NxM_y=N_x. These are logically independent properties: some equations are separable but not exact, some are exact but not separable (such as Example 2 above, where exsin⁡ye^x\sin y mixes the variables inseparably), and some are both or neither. Exactness is checked by differentiating MM and NN; separability is checked by attempting to factor the equation into single-variable pieces.

What if I find FF but forget to add the arbitrary constant CC?

The differential equation only pins down FF up to an additive constant, since d(F+c)=dFd(F+c) = dF for any constant cc. The actual content of the solution is the family of level curves F(x,y)=CF(x,y)=C, one curve for each value of CC; an initial condition selects the specific member of the family. Writing "F(x,y)=0F(x,y)=0" instead of "F(x,y)=CF(x,y)=C" silently throws away every solution except one, so the general solution must always retain the arbitrary constant until an initial or boundary condition is imposed.

Does the potential function always come out algebraically nice, like a polynomial?

No — FF can involve transcendental functions, as in Example 2 where F(x,y)=exsin⁡y+2ycos⁡xF(x,y)=e^x\sin y + 2y\cos x. The construction algorithm (integrate MM in xx, differentiate to solve for the missing piece in yy) works regardless of the functional form of MM and NN, as long as the relevant antiderivatives can be computed. If the required antiderivative has no elementary closed form, FF may need to be left as a definite integral, though the implicit solution F(x,y)=CF(x,y)=C remains perfectly valid and well-defined.

What happens if My≠NxM_y \neq N_x — is the equation unsolvable?

Not necessarily; it means the equation is not exact as currently written, but it may become exact after multiplying through by a suitable nonzero function μ(x,y)\mu(x,y), called an Integrating Factor. The product μM dx+μN dy=0\mu M\,dx + \mu N\,dy = 0 has the same solution curves as the original equation (since multiplying by a nonzero factor does not change where the differential vanishes), but can satisfy (μM)y=(μN)x(\mu M)_y = (\mu N)_x even when the original M,NM,N did not. Finding such a μ\mu is the natural next technique after exactness.

Why does simple connectedness matter, in plain terms?

Simple connectedness means the domain has no holes that a loop could get "stuck around." The proof of sufficiency builds FF by integrating along paths from a fixed base point, and for FF to be a well-defined function, every path between the same two points must give the same answer. Green's theorem guarantees this path-independence, but only when every closed loop in the domain bounds a region entirely contained in the domain — exactly the definition of simple connectedness. The punctured-plane counterexample shows that when a loop encircles a missing point, the integral around it can be nonzero even though My=NxM_y=N_x holds everywhere it is defined.

Can an exact equation have singular or extra solutions not captured by F(x,y)=CF(x,y)=C?

Generically no — every differentiable solution curve of an exact equation with continuous M,NM,N satisfies F(x,y)=CF(x,y)=C for some constant, by the direct chain-rule argument in the definition. However, care is needed where N(x,y)=0N(x,y)=0, since the equation may then fail to be solvable for yy as a function of xx (the solution curve could have a vertical tangent there); the implicit relation F(x,y)=CF(x,y)=C still correctly describes the curve, even at such points, while an explicit y=ϕ(x)y=\phi(x) form might not exist locally.

Is testing My=NxM_y=N_x enough, or should I also check N,M∈C1N,M \in C^1?

Both matter. The theorem explicitly requires MM and NN to have continuous partial derivatives My,NxM_y,N_x on the domain in question; this is what makes Clairaut's theorem and the differentiation-under-the-integral steps valid. If MM or NN has a jump discontinuity or the partials fail to exist at some points of the domain, the equality My=NxM_y=N_x elsewhere does not guarantee a global potential function exists, and the construction in the proof can break down at the bad points.

Before this