Differential Equations

By Michael on April 17, 2008 at 1:09 am | In Blog Posts |

I’m taking a class on Differential Equations. These are hard to explain but I’m going to give it a try. To do so I am going to try to explain differential calculus in a nutshell.

Your driving down the road in an automobile. You have an odometer in the car which tells you how far you’ve gone (x). You have a speedometer which tells you how fast you are going (v). When you punch the gas pedal or hit the brakes, your car accelerates (a), getting faster or slower. All of these quantities are related through calculus. Let’s say you start at x=0 and drive at v=40 miles per hour for t=15 minutes.

x = vt
x = (40mph*0.25 hours)=10 miles

We can write this as x/t = v. The velocity is the ratio of the change in position (x) to the change in time (t).

In math instead of saying “change in x” we say “delta-x” . In calculus we take this delta to be infinitely small and instead of saying “delta-x” we say “dee-x” (and write dx). So when I write dx I just mean a ridiculously small change in position. dt means an instantaneously small slice of time. The equation v=x/t can be written with this notation as v=dx/dt. This means the same thing as above, it’s the same definition of velocity, but we are implying these infinitely small deltas.

Another way of saying dx/dt is to say “the time derivative of x”. This just means “how x changes with time”. If we assume that we are talking about time we can use a shorthand which means “how x changes with time” and that shorthand is simply a single apostrophe like this: x’.

Now we can write our equation above like this:

v = x’

But your velocity can change over time, too. If you are going 40 mph and accelerate to 60 mph, your velocity changes as a function of time.

v = at

For some length of time (t) you accelerated at a rate (a) and your new velocity is your original velocity plus (a) times (t). If we rewrite:

a = v/t

and using our notion of infinitely small deltas:

a = dv/dt

we find that (a) has the same relationship to (v) that (v) had to (x) and thus

a = v’ = x”

So we have:

x The position
x’ The velocity
x” The acceleration

The velocity is referred to as the “first derivative of x” and the acceleration is referred to as the “second derivative of x”. These derivatives have a mathematical relationship that is beyond the scope of this article.

Now to differential equations. These are equations where we see complicated relationships between values and their derivatives. An example is a falling object (m) reaching terminal velocity due to wind resistance. The force on this object (F=ma) is the difference between the gravity (g) of Earth pulling it down (-mg) and the wind resistance (k) getting stronger as the object falls faster (kv):

ma = kv - mg

For simlicity we divide by m and just let k = k/m:

a = kv - g

and now use our fancy notation from above:

x” = kx’ - g

or equivalently:

v’ = kv - g

In this equation the change in velocity is a function of the velocity itself. So to know the acceleration you need to know the velocity but to know the velocity you need to know the acceleration!

This is the conundrum of differential equations. Finding functions that, when you find their derivative, give themselves back.

If you solve the equation v’ = kv - g the answer you get is the velocity as a function of time. If you plug in a very long time, the velocity reaches a limiting value — the terminal velocity.

You can’t really solve this problem without differential equations. It turns out there are many, many things in the real world, in physics, biology and engineering, that can be explained with differential equations. From orbits to population growth to thermodynamics, differential equations are actually quite useful.

Congratulations if you stuck with me! If you guys are interested we can talk a little about differential equations in astronomy next time.