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The chain rule gives $dF(x,y(x))/dx=F_x+F_y{y}^{\prime }$. Equality with $P+Q{y}^{\prime }$ for every $y(x)$ therefore requires $F_x=P$ and $F_y=Q$, which implies $P_y=Q_x$. Conversely, on the plane this compatibility condition makes $Pdx+Qdy$ an exact differential, so a potential $F$ exists.

Here $P=4x^3+3y$, $Q=2y+3x$, and $P_y=Q_x=3$. Integrating gives $F=x^4+3xy+y^2$, hence

On the simply connected domain ${\mathbb{R}}^3$, a continuously differentiable differential one-form is exact if its mixed partial derivatives agree, equivalently if its associated vector field has zero curl. Here one can verify exactness directly by observing thatsatisfiesThus $udx+vdy+wdz=dF$. By the fundamental theorem for line integrals, its integral is path independent and equalsConsequently, for every path with the stated endpoints,

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.

Here $M_y=2y$ whereas $N_x=0$, so the equation is not exact. The identity $M_y-N_x=N$ gives the integrating factor for a differential one-form $\mu (x)=e^x$. Multiplying makes the left-hand side the exact differential of $e^x(y^2-x+1)$. ThusFor real graph solutions, the right-hand side must be nonnegative; a smooth choice of square-root branch is made on the interval concerned. Finding the factor did not require dividing by $y$ at zero.

The cross derivatives are both $2y$, so the equation is exact. The function $x\cos x+x{y}^2$ differentiates to the given one-form. HenceDifferentiating this implicit equation checks both coefficients of the exact differential directly.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

The product rule givesThus this is an exact differential, and its zero level differential means $y{\sin }^{2}x=C$. The initial data set $C=1$, givingThe displayed interval is the maximal connected real interval containing $\pi /2$ on which this finite-valued solution is defined.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. Here this givesOn an open two-dimensional domain the coefficient relation is $x{\mu }^{\prime }+3\mu =0$, so $\mu =C{x}^{-3}$ on a connected interval avoiding zero. The normalization gives $C=1$ on the positive interval through $x=1$. Integrating $f_x=2x^{-3}\sin y$ and checking $f_y$ yieldsThere is no smooth extension of this normalized integrating factor across $x=0$; an independently specified factor on $x<0$ is not fixed by data at one.

If $Pdx+Qdy=dF$ with continuous second partial derivatives of $F$, then $P=F_x$, $Q=F_y$. The symmetry of second derivatives requiresThis is a necessary condition for an exact differential. With continuously differentiable coefficients it is also locally sufficient, and is globally sufficient on a simply connected open domain; a domain with holes can require additional global information.

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

Let $M=x+y$ and $N=x$. Their cross partial derivatives agree: $M_y=N_x=1$. More directly, construct the scalar functionThis exhibits the expression as an exact differential. Along a solution of the differential equation,Thus the general solution isFor $C=0$, the graph $y=-x/2$ also extends smoothly through $x=0$ and satisfies the original, undivided equation there. Nonzero $C$ cannot extend through zero as a finite-valued differentiable solution. The extra vertical component $x=0$ of the implicit zero-level equation is not a graph $y(x)$ on an interval and is not an additional solution of this ordinary differential equation.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.

When the Hamiltonians agree, exactness of the one-form givesThe action argument of part b then proves canonicity. Equivalently, exterior differentiation preserves the symplectic two-form. For a time-dependent transformation, the exact differential must supply the full action identity; a fixed-time check alone would not establish that the new Hamiltonian equals the old one without a time-dependent correction.

A line integral is path independent when its value depends only on the two endpoints. On a connected domain, this holds for the line integral of every exact differential.

For a vector-field line integral, this condition is equivalent to the field being a conservative vector field.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.