How do we go about solving this equation? Well, the first thing that we're going to

do is look at it and think. dx dt equals a times x.

Let's simplify and say that a equals 1. We're looking for a function whose

derivative is itself. You and I both know an excellent

solution, that is x of t is e to the t. Now, what happens when we put the a back

in there? Well, if we just think about it a little

bit more, we see that putting an a up in the exponent gives us, as its derivative,

e to the a t times a. That is what we're looking for.

We could, of course, put an arbitrary constant, C out in front.

And you may check that this is a solution to that differential equation.

What is that constant? Well, we might, following our previous

example, call it x not, the initial condition.

What you get at time 0. This method of solution is sometimes

called solution by ansatz. Which is a fancy word for saying, we took

a guess, and it turned out to be correct. There is, of course, a more principled

approach. One such approach is based on series

where we assume that our function x has a series expansion c0 plus c1t plus c2 t

squared, etcetera. These constants c sub k are, to us,

unknown. Now, what are we going to do with this?

Well, the differential equation tells us something about a derivative of x with

respect to t. We can differentiate the terms of this

series very easily. What the differential equation tells us

is that this is equal to a times x, which of course is a times c0, plus a times

c1t, plus a times c2 t squared, etcetera. Now, here is the important step.

If we have two series that are equal than their coefficient in front of the various

terms must also be equal, and now we can work one step at a time.

The constant terms equating them, tell us that c1 is equal to a times c0.

Well that's good. If we knew c0 we would know c1.

What does the next equation, that of the first order coefficients tell us?

Well, it tells us that c2 is equal to 1 half a times c1.

But we already know what c1 is, c1 is a times c0, so that tells us that c2 is

actually 1 half a squared times c0. What happens when we take the

coefficients of the quadratic term and set them equal to each other?

Well, 3 times c3 equals a times c2. That means c3 is 1 3rd a c2.

Using what we know about c2, we get 1 over 3 factorial.

Times a cubed, c not. I'm going to let you keep going with

this, and see if you can find a pattern. That pattern through the method of

induction, can show that c sub k is 1 over k times a times c sub k minus 1.

And that is, 1 over k factorial, a to the k times C not.

That means, when we look at our original series, x of t, it's the sum over k of

ck, t to the k, that is c0 times 1 over k factorial times a to the k times t to the

k. Now,, I'll let you verify that, that is

really c not times the exponential function e to at.

We obtain our solution, which we already knew was true, but from a more

principled, serious approach. That however, is a little complicated,

and so we are led to our last and best method of solution, that of integration.

Beginning with x as a function of t, we differentiate using the chain rule to

obtain dx is dx dt times dt. Now, we notice something about dx dt

since this is a solution to the differential equation.

It is a times x. And now, we're going to use the method of

separation, which means we move all of the x terms over to one side of the

equation, and we keep the, t terms, over to the other side.

This gives us dx over x equals adt. And now, using the fact that the integral

is an operator, we're going to apply it to both sides of this equation.

Obtaining the integral of dx over x equals the integral adt.

What is the integral of dx over x? Well, it's the anti-derivative.

Natural log of x. What is the integral of a constant times

dt? It's simply at plus an integration

constant. Now, to solve for x, we apply the

exponentiation operator. E to the log of x is simply x.

On the right hand side, we get e to the at plus a constant.

Splitting that up into a product, we obtain a new constant.

And we'll call that c as well, times e to the at.

That is our solution and it was obtained effortlessly.