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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.

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This question has an open bounty worth +50 reputation from JohnD ending in 5 days.

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On the one year anniversary of this question, let's bring more attention to it with a bounty and find an answer (if there is one)!

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Sorry, it is the TWO YEAR ANNIVERSARY of the question, not one year! –  JohnD 2 days ago

4 Answers 4

99.9% of equations in the real world are ugly, nonlinear and generally unsolvable by direct analytic means. What we CAN do is for example, look at linearizations of such equations.

For example, even an equation for a swinging pendulum is only solvable through elliptic functions, whereas the linearizion looks something like $\ddot{\theta}=-\frac{g}{L}\theta$ which is nice and solvable. This would be the first order approximation, and is generally "valid" for small movements about the origin. Why is this useful? During the moon landing through video, NASA scientists noticed that the American flag (or maybe it was a small tassel atop it) was rocking lightly about the stick it was on. From the linearization of the pendulum equation one can quickly derive that the period of oscillation is $T=\frac{1}{2\pi}\sqrt{g/L}$ where $g$ is the gravity and $L$ is the length of the pendulum. You can easily solve this for $g$ so that the scientists were able to get a rough estimate for the strength of gravity on the moon just from a grainy black-and-white video. Not entirely dissimilar methods were used by scientists like Huygens and Galileo hundreds of years ago to measure the earth's gravity.

Otherwise, the local behavior of many nonlinear systems can again be studied by linearizations which remain valid for small movements. The sad fact is that the only equations for which we have $\it{immediate}$ and simple analytic solutions are in fact linear equations. I would go so far as to say (and I believe Feynmann did as well) that the pendulum equation above is essentially the bread and butter of all of physics in that it's almost not unreasonable to say that it's one of the few equations we can actually solve truly explicitly (and still be useful!). Even in quantum mechanics and quantum field theory, oscillations to first order essentially obey this simple second order linear equation. To put it bluntly: when faced with something that is highly nonlinear and complicated, your first guess should be whether or not approximations can be made to see linearized behavior.

As far as exact ODE's go, yes for practical purposes exact solutions are definitely useful. Linearizations behave terribly away from the points which they are calculated and moreover phenomena is lost upon linearizations. Any sort of interesting nonlinear behavior is precisely the consequence of nonlinearity and cannot be realized in any simple linear system. Moreover, exact ODE's can have high order approximations. For example, a complicated integral can be numerically approximated. If you're building an airplane then your tests in wind tunnels will require you to consider exact ODE's because the complex interaction of everything involved is completely washed away if one takes a simpler approach.

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I am certainly aware that textbook problems are just that---textbook problems---and don't expect them to be highly precise models of complicated phenomena, devoid of simplifying assumptions. The point of my question, however, is more along these lines: are exact equations as "useful" (in these sense of solving the textbook type applications listed above) as their separable and linear siblings? I cannot think of an example where they are. (Note: I have absolutely no problem with studying equations that have zero physical relevance. But that's not my question here.) –  JohnD Dec 19 '12 at 22:33
    
@JohnD: I'm not sure what you mean by your distinction between seperable and nonseperable problems. Just because your differential equation is seperable doesn't mean that you can solve it analytically (or rather, easily for that matter). And usefulness is completely dependent on what you are looking for. If you are studying local behavior, linearization suffices for many inferences that you could want. The equation $x''=-c\sin(x)$ which is the full pendulum equation is seperable yet solvable only by elliptic functions and is not trivial. –  Alex R. Dec 19 '12 at 22:58
    
@JohnD: and naturally if you wanted to study again, say a pendulum then for large oscillations using the full elliptic function formalism is unavoidable so depending on the exactness of what you need, of course exact ODE's are quite useful for precision. –  Alex R. Dec 19 '12 at 23:04
    
Just to reiterate: It seems you are using the word "exact" to be "precise" or "accurate", whereas I am using it as the class of DEs of a particular type. –  JohnD 52 mins ago

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.

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I believe that the question being asked is not about exact solutions to differential equations but a certain class of differential equations which are termed (exact differential equations). Two of the reason that they are useful is because there is a known method for solving them, and exact equations include a certain class of non-linear equations which are unsolvable by the methods you've learned for separable and linear equations. The more techniques you know and the more types of equations you know how to message information out of the more useful you will find differential equations for studying the real world (or for understanding pure mathematics). Hope this was helpful.

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I am looking for practical, physical phenomena which are modeled by exact ODEs (but not any of the other standard classes of ODEs). This is my particular sense of "useful". –  JohnD Feb 22 at 22:34

Wikipedia references:

<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.

enter image description here

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.

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+1 This is the kind of answer/application I am after. –  JohnD 51 mins ago

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