This comment was posted to reddit on Mar 23, 2015 at 5:12 pm and was deleted within 10 hour(s) and 49 minutes.

Well, the simplest differential equation is the one where the rate of change of something depends on its quantity. So for example, a fixed rate savings account at a bank accrues interest depending on its instantaneous principle. Of course real banks do this discretely, but they could do this continuously (which is the version we are interested in). How does one solve this? You must first set up an equation:

df(x)/dx = (1+z)f(x)

Where z is your instantaneous interest rate. If you have already seen diff-eq's in class, then you know that this is easily solvable via the substitution: f(x)=exp(a*x)g(x), at which point you quickly learn that a = (1+z), and g(x) = constant = f(0).

Same principle applies to radiometric decay. Suppose atoms of uranium decays to lead because of uniformly random cosmic ray bombardments. So if you have a big lump of uranium, its rate of decay to lead is proportional to the amount of uranium you have at any one time. So:

df(x)/dx = -l*f(x)

which is quite similar to the interest equation above, and solves via the same substitution to f(x) = f(0)*exp(-l*x).

So these are really easy differential equations. But what happens when it is much more complicated, like Uranium, U(t), decaying to Thorium, but then the Thorium, T(t), itself is decaying to Rubidium, R(t), at some rate. Well U(t) is known and solvable from the above techniques. Suppose we started with some amount of thorium and just watched it decay to rubidium, then we would just have:

dT(t)/dt = -r*T(t)

But the rate of thorium change in our situations is equal to the amount of decay from uranium minus the amount which decays to rubidium:

dT(t)/dt = -dU(t)/dt - r*T(t)

Well, ok, the U(t) formula, we know is basically U(0)*exp(-u*t) so we are left with:

dT(t)/dt = u*U(0)*exp(-u*t) - r*T(t)

This is a little more complicated (an exercise to the reader). Basically by having a relationship between the rate of something and its amount, we have a differential equation.