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For an exact differential form, equality of the mixed partial derivatives requires $P_y=Q_x$. HereEquality for all $x>0$ and real $y$ forces $1/a=1/2$, so $\boxed{a=2}$. To construct the potential, integrate $P$ with respect to $x$:Differentiation with respect to $y$ gives $v_y=Q+C^{\prime }(y)$, so $C^{\prime }(y)=0$. Thus the most general potential on the connected domain $x>0$ isIts explicit partial derivatives verify sufficiency as well as necessity.

Let $P=\mu (x)(xy-16x-x^3)$ and $Q=\mu (x)(16+x^2)$. An exact differential form must satisfy $P_y=Q_x$. These partial derivatives giveTaking antiderivatives produces $\ln |\mu |=-\frac{1}{2}\ln (16+x^2)+C$. A convenient nonzero integrating factor for a differential one-form is thereforeAn explicit potential, which also proves global exactness, isIndeed $F_x=P$ and $F_y=Q$ for the chosen integrating factor.

For the given differential form, set $P=x(y^4+1)$ and $Q=y(2x^2{y}^2+1)$. Their partial derivatives satisfyIn fact an explicit potential issince $F_x=P$ and $F_y=Q$. Thus the differential form is an exact differential form on all of ${\mathbb{R}}^2$.

Replacing $y^2$ by $u$ suggests the potentialDirect differentiation verifies that $dH/dx$ is exactly the left-hand side of the required ordinary differential equation. This direct verification remains valid even for negative $u$; it does not require a real square root $y=\sqrt{u}$. ThereforeThe initial condition gives $C=1$. Solving this quadratic equation in $u$ and selecting the branch that has $u(1)=0$ givesThe second expression also defines the removable value $u(0)=1$. The maximal open interval containing $1$ on which this is a differentiable real solution isAt either endpoint the square root vanishes and $2x^{2}u+1=0$, while $x(u^2+1)\ne 0$, so the original ordinary differential equation cannot hold there with a finite derivative.

Every smooth closed differential form of positive degree is locally an exact differential form. On a coordinate ball centered at zero, put $F_t(x)=tx$ and let $R=x^i{\partial }_i$ be the radial vector field. Define the degree-lowering operatorFor a $p$-form with $p>0$, the integrand has a factor $t^{p-1}$ in its coordinate coefficients and is smooth at zero. Cartan's magic formula and differentiation of the pullback of a differential form give ${\partial }_t(F_t^{\ast }\alpha )=t^{-1}{F}_t^{\ast }(d{\iota }_R\alpha +{\iota }_{R}d\alpha )$. Integrating from zero to one gives $dK\alpha +Kd\alpha =\alpha$, since $F_0^{\ast }\alpha =0$ in positive degree. If $d\alpha =0$, take $\beta =K\alpha$. The local potential of a closed differential one-form gives the explicit degree-one construction. Global exactness can fail through nonzero periods, measured by de Rham cohomology.

For a Lagrangian submanifold $L$ in an exact symplectic manifold with primitive $\lambda$, its period subgroup is $\{{\int }_c\lambda :c\in H_1(L;\mathbb{Z})\}\subset \mathbb{R}$. It is rational when this subgroup is discrete. If it is nonzero, it is $\gamma \mathbb{Z}$ for a unique $\gamma >0$. Some conventions reserve the word rational for this nonzero case. The zero subgroup corresponds to an exact differential form $\lambda {|}_L$; its least positive period is interpreted as $+\infty$ if that endpoint convention is admitted.

On a nonempty compact $2n$-dimensional symplectic manifold without boundary, with $n\ge 1$, the class $[\omega {]}^n$ in top-degree de Rham cohomology is nonzero, since its integral in the symplectic orientation is positive. In particular $\omega$ cannot be an exact differential form: if $\omega =d\alpha$, then ${\omega }^n=d(\alpha \wedge {\omega }^{n-1})$ and the Generalized Stokes theorem would make that integral zero. Thus vanishing second de Rham cohomology obstructs a closed positive-dimensional symplectic manifold.

The transform integrates a base function and a velocity-evaluated one-form along the closed trajectories of a surface thermostat. Every exact differential form of degree one integrates to zero, because $F(f\circ \pi )=df(v)$. The relevant kernel theorem concerns these functions of velocity degree zero and one, rather than arbitrary functions on the unit tangent bundle.

Let $n\ge 1$. For the double covering map $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ and the antipodal map $a$, every pulled-back top differential form satisfies $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$. Since $\mathrm{deg}a=-1$, its integral over the sphere is its own negative and hence vanishes. Top-dimensional de Rham cohomology of the oriented sphere therefore makes it an exact differential form.

Average a primitive under $a$ to obtain an invariant primitive. It descends through the double covering map, proving that $\omega$ is exact on Real projective space. This is the invariant primitive under a finite group action argument. The condition $n\ge 1$ is essential: ${\mathbb{RP}}^0$ is a point, whose degree-zero de Rham cohomology is $\mathbb{R}$.