Practical applications of first order exact ODE?

In elementary ODE textbooks, an early chapter is usually dedicated to first order equations. It is very common to see individual sections dedicated to separable equations, exact equations, and general first order linear equations (solved via an integrating factor), not necessarily in that order.

Common practical applications in these texts include population growth/decay, mixing problems, draining tank/Torricelli's Law problems, projectile motion, Newton's Law of Cooling, orthogonal trajectories, melting snowball type problems, certain basic circuits, growth of an annuity, and logistic population models. (This is just off the top of my head so maybe I am missing other popular ones.) However, all of these end up as separable or first order linear problems and are solved accordingly.

Are there practical applications that lead to first order ODEs which are (exclusively) exact equations?

Edit: To clarify, I am not saying that exact equations are never useful. I am simply inquiring about their relevance/applicability in the very particular context mentioned above.

To put the question another way, can you briefly state (e.g., in the form of an exercise that would appear in popular undergrad ODE books like Boyce & DePrima; Zill; Nagle/Saff/Snider; Edwards & Penney; etc.) an application problem modeled by a first order exact ODE (which is not separable or linear) and that is solvable by hand? I've looked in the dozen or so ODE textbooks on my shelf and none of them contain such a problem. I find that absence curious.

Wikipedia references:

• Streamlines, streaklines, and pathlines
• Stream function

<quote> Streamlines are a family of curves that are instantaneously tangent to the velocity vector of the flow. These show the direction a massless fluid element will travel in at any point in time. </quote>
Consider the velocity field $(u,v)$ of a two-dimensional incompressible flow. Let the family of curves be given by $\;\psi(x,y) = c$ . The velocity vectors are tangent to these as shown for one of them in the following picture.

Thus, along the curve $\psi(x,y) = c$ , the following equations hold: $$ \left. \begin{array}{l} \frac{dy}{dx} = \frac{v}{u} \\ d\psi = 0 = \frac{\partial \psi}{\partial x} dx + \frac{\partial \psi}{\partial y} dy \end{array} \right\} \qquad \Longrightarrow \qquad \frac{dy}{dx} = - \frac{\partial \psi / \partial x}{\partial \psi / \partial y} = \frac{v}{u} $$ Hence, apart from a constant: $$ u = \frac{\partial \psi}{\partial y} \qquad ; \qquad v = - \frac{\partial \psi}{\partial x} $$ But the flow is incompressible, so: $$ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \qquad \Longrightarrow \qquad \frac{\partial^2 \psi}{\partial x\, \partial y} = \frac{\partial^2 \psi}{\partial y\, \partial x} $$ Herewith the conditions for an exact differential equation are fulfilled. Now solve $\psi$ from: $$ v\, dx - u\, dy = 0 $$ Example. As taken from : Find the velocity of a flow . $$ u = -\frac{y}{x^2+y^2} \qquad ; \qquad v = \frac{x}{x^2+y^2} $$ Then: $$ v\, dx - u\, dy = \frac{x\,dx + y\,dy}{x^2+y^2} = \frac{d\left( x^2+y^2 \right)}{x^2+y^2} = 0 \qquad \Longrightarrow \qquad x^2 + y^2 = c $$ It is concluded that the streamlines of this flow are circles.

Example. Somewhat related to the above one. $$ u = \lambda\,x \qquad ; \qquad v = \lambda\,y $$ Then, assuming that $\; x\ne 0$ (i.e. $\,x=0\,$ as a special case) : $$ v\, dx - u\, dy = 0 \quad \Longleftrightarrow \quad \frac{y\,dx - x\,dy}{x^2} = - d(y/x) = 0 \quad \Longrightarrow \quad y = c\, x $$ An integrating factor has been used. It is concluded that the streamlines of this flow are straight lines through the origin.

Wikipedia reference:

• Electric potential

The electric potential at a point $\vec{r}$ in a two-dimensional static electric field $\vec{E}$ is given by the line integral: $$ V = - \int_C \vec{E}\cdot d\vec{r} = - \int_C \left(E_x\, dx + E_y\, dy\right) $$ where $C$ is an arbitrary path connecting the point with zero potential to $\vec{r}$. It follows that: $$ E_x = - \frac{\partial V}{\partial x} \qquad ; \qquad E_y = - \frac{\partial V}{\partial y} $$ The integral is zero if the path is closed. Then Green's theorem tells us: $$ \oint \left( E_x\, dx + E_y\, dy \right) = \iint \left( \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right) dx\,dy = - \iint \left( \frac{\partial^2 V}{\partial x \, \partial y} - \frac{\partial^2 V}{\partial y \, \partial x}\right) dx\,dy = 0 $$ Thus establishing once more the conditions for solvability of the exact differential equation: $$ E_x \, dx + E_y \, dy = 0 $$ Solving this ODE results in the iso-lines $\,V(x,y) = c\,$ of the electric potential $\,V$ .

Example. An infinitely long and infinitely thin charged wire perpendicular to the plane and intersecting it in the origin. Apart from constants: $$ (E_x,E_y) = \frac{(x,y)}{r^2} \quad \Longrightarrow \quad E_x\, dx + E_y\, dy = \frac{x\,dx + y\,dy}{r^2} = 0 \quad \Longrightarrow \quad x^2+y^2 = c $$ The equipotential lines are circles.
Can of worms: special cases, and a singularity at the origin in all of the examples.

I am not quite sure that this is exactly an example you are looking for, but still.

Consider the Lotka--Volterra system on the plane $$ \dot x=x(a_1x+b_1y+c_1)=P(x,y),\\ \dot y=y(a_2x+b_2y+c_2)=Q(x,y).\tag{1} $$

The following theorem is true: System $(1)$ does not have limit cycles. The proof is based on the Dulac's criterion that the expression $$ \frac{\partial}{\partial x}(BP)+\frac{\partial}{\partial y}(BQ)\tag{2} $$ has a definite sign. Here $B=x^{k-1}y^{h-1}$ is an integrating factor and $k,h$ depend on the parameters of the model.

It is possible that expression $(2)$ can be zero, in this case (and this is your example) the equation $$ \frac{dy}{dx}=\frac{Q(x,y)}{P(x,y)} $$ admits the integration factor $B$ and, after multiplication by $B$, is exact, hence admitting an analytic integral. And in this case therefore the phase plane consists of closed orbits.