# Velocity-Dependent Resistive forces

1. May 25, 2013

### chief114

1. The problem statement, all variables and given/known data

Could someone explain me/show how to integrate the following differential equation regarding velocity and resistance force.

2. Relevant equations

differential equation: dv/dt=g-(b/m)v.

The resulting equation comes out to be: v=(mg/b)(1-e^-bt/m).

3. The attempt at a solution

2. May 25, 2013

### rude man

Separation of parts.

Rearrange the equation so the left side has only v and possible constants(s) and the right only t and possible constants.

Then integrate both sides. Don't forget the constant of integration!

3. May 25, 2013

### cepheid

Staff Emeritus
Welcome to PF chief114!

First of all, you don't "integrate" a differential equation. You "solve" it. This may seem like semantics, but it's an important distinction. The whole point of differential equations (what makes them different from simple differentiation or integration problems) is that you don't *know* the function v(t), so you have nothing to integrate. All you know is that there is some relationship between v(t) and its first derivative. Solving the differential equation means finding a function v(t) that satisfies this relationship (i.e. that makes it true).

Now, on to answering your actual question: to solve a differential equation of this type, I like to use the method of integrating factors. An integrating factor is something that you can multiply both sides of the equation by, and that makes the left hand side an exact differential (i.e. makes it something that is the derivative of something else). Once the left hand side is an exact differential, you *can* just integrate it in order to solve the equation. The details of the method are here:

http://en.wikipedia.org/wiki/Integrating_factor

4. May 25, 2013

### cepheid

Staff Emeritus
This equation is not separable, as far as I can see.

5. May 25, 2013

### barryj

When I take the derivative of v=(mg/b)(1-e^-bt/m) I don't get dv/dt=g-(b/m)v do I?
Is there a typo somewhere?

6. May 25, 2013

### cepheid

Staff Emeritus
The answer given is correct provided v(0) = 0.

7. May 25, 2013

### rude man

dv/[g - (b/m)]v = dt

Always use separation if possible. So easy!

8. May 25, 2013

### cepheid

Staff Emeritus
It's not dv/[g - (b/m)]v = dt

It is: dv/[g - (b/m)v] = dt

The v doesn't distribute through to the g term, so you can't factor it out from it.

Can you show me your method actually leading to the correct answer? Inhomogeneous first order differential equations aren't separable, from what I can remember.

9. May 25, 2013

### rude man

It is: dv/[g - (b/m)v] = dt
[/QUOTE]

So it is. Here we go:

dv/[g - (b/m)v] = dt

Using ∫dv/(g-av) = - [ln(g - av)]/a where I am abbreviating a = b/m

we get

- [ln(g - av)]/a = t + constant

which after some algebra leads to

v = (1/a)(g - K*exp(-at)
with K a constant (units of acceleration) to be determined by initial conditions.

If v(0) = 0 then K = g and we get the OP's equation.

10. May 26, 2013

### cepheid

Staff Emeritus
Interesting. I have always solved these types of equations* using the method of integrating factors. I think I still prefer it because it has a certain elegance. Multiply both sides by exp[(b/m)t] and the left hand side becomes an exact differential. So integrating it involves doing nothing at all. The right hand side is a constant times an exponential function, which is trivial to integrate. BOOM! You're done. Your method requires knowing what the integral of (g-av)-1 is, or looking it up, or working it out, which is fine, but seems marginally more work.

*When I say "these types of equations", I mean the INhomogeneous ones. Obvously v' + Cv = 0 is trivially separable. I think the reason is that I learned method of integrating factors for other inhomogeneous equations where you can't use your trick, because the coefficients are functions of t, instead of being constant, and I was remembering that more general case.

Let's see if you really know what you're doing. Pop quiz. So you had something of the form dv/f(v) = dt, and you "integrated both sides." But when you really think about it, you integrated the left-hand side with respect to v, and the right hand side with respect to t, which is wrong, because it involves not doing the same thing to both sides. Why does this work?

11. May 26, 2013

### rude man

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) ...

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."

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.

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.

Last edited: May 26, 2013
12. May 26, 2013

### cepheid

Staff Emeritus
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.

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.

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.

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.

13. May 26, 2013

### rude man

You seem to have taken that personally. Not so intended.

I based my statement on a very large number of physicists, and more than a few mathematicians also, with whom I've worked over lo these many years. They were hot stuff with the Fourier transform but did not know the Laplace.