rude man said:
Oh you mentors! I thought I was through with pop quizzes when I gaduated from college!
OK, here we go:
any 1st order ode of the form
M(x) + N(y)dy/dx = 0
can be separated.
We rewrite this as
M(x)dx + N(y)dy = 0 = M(x)dx + N(y)dy/dx dx
Integrate both sides wrt x:
∫M(x)dx + ∫N(y)dy/dx dx = constant
But the second integral is also ∫N(y)dy
permitting
∫M(x) dx + ∫N(y)dy = constant
(See, I know how to look things up in
Thomas ...)
I would add that integrating ∫dv/(g - av) is pretty darned easy. Let u = g-av, du = -a dv, and most of us know that ∫du/u = ln(u) ...
I'm sure the above is all fine. But what I was actually going for was the following: instead of writing the equation we had above in the form f(v)dv = dt, (where f(v)=1/(g-av)) you instead should really write this:
$${f(v)}\frac{dv}{dt} = 1$$
Usually you can think of f(v) as being the derivative of something, i.e. f(v) = dF/dv, where F(v) is an antiderivative of f(v). So this becomes:
$$\frac{dF}{dv}\frac{dv}{dt} = 1$$
which, according to the chain rule, becomes:
$$\frac{d}{dt}\left[F(v)\right] = 1$$
Now we can integrate BOTH sides with respect to TIME (not one side with respect to v and the other with respect to time). And since F(v) = ∫f(v)dv by definition, this yields:
$$F(v) = \int f(v)\,dv = \int dt$$
THAT (i.e. this property of the chain rule) is why what you did above works. Doing it this way avoids this business of separating dv/dt into dv and dt, which is playing fast and loose with notation.
rude man said:
As for the alleged superioprity of the integrating factor method, let me quote Thomas:
"Unfortunately there is no general technique for finding an integrating factor when you need one, and the search for one can be a frustrating experience."
Actually, for this specific type of differential equation: a first-order, linear, inhomogeneous ordinary differential equation:
dv/dt + μ(t)v(t) = g(t)
he integrating factor is always just exp(∫μ(t)dt) where μ(t) is the coefficient on the v term. You can prove this relatively easily just by searching for a way to make the lefthand side of the above equation an exact differential. In this case, since μ(t) was just a constant b/m, I immediately knew that the integrating factor was exp[(b/m)t]. I did not have to think about it
at all.
rude man said:
When I taught a diff eq course (the less said about this the better ...) I decided to omit this method altogether as being of marginal value.
How unfortunate for you. This technique is extremely useful, especially considering how often a DE of this form comes up, in everything from terminal velocity problems to RC circuits.
rude man said:
Of course, as an EE I do all these equations with the Laplace transform, which is by far the best method as I have argued in these hallowed pages before. All initial conditions and forcing functions automatically included. No "guessing" solutions. But alas, mathematicians and physicists seem to have an elitist aversion to using the tables, if they know the method at all, which most of them don't.
I took a program that combined EE + physics, and I learned Laplace transforms both in the context of a practical engineering course, and more rigorously in an applied math class. Don't assume things about people who you don't actually know.
