6.4.4

Instructor: In example 4, we want to use power series to solve the initial value problem y prime equals y, with y of 0 equals 3. We've actually seen this problem once before, I believe. But we're going to use a power series to solve this. Technique will be a little different. So first, we want to say that y is a power series itself, and just give it coefficients c0 plus c1x plus c2x squared plus c3x cubed plus and off to infinity. Well that means y prime is equal to c1 plus 2c2 plus 3c3 plus 4c4 plus, and off to infinity. Oh, those are just my coefficients. How about x there, x squared, x cubed, et cetera. 

Now since we know y prime equals y, that means that we have a series of facts. c0 must be equal to c1. c1 must be equal to 2c2. c2 must be equal to 3c3, and so forth. So solving this, we know that c0 is equal to 3. And because we have our y of 0 equals 3, that would mean that if we evaluate that at 0, we can c0 equals 3, OK? So c0 equals c1 equals 3, which it will be useful to know that is 3/1 factorial. 

Well, c2 is a half of that. It's 3/2, which is 3/2 factorial. Factorials come up an awful lot with power series, as we've seen. c3 is 3 divided by-- well actually it's 3/2 divided by that, right? Because c3 would be c2, which is 3/2 factorial divided by 3, which is 3/3 factorial, and so forth. 

So that looks to vary by n's. Well, let's evaluate this into our function y equals, that we claim to be a power series. 3/1 factorial plus 3/1 factorial x plus 3/2 factorial x squared, plus 3/3 factorial x cubed, plus 3/4 factorial x fourth, continuing that pattern. Well if you notice, we could actually factor a 3 out of this to be 1 plus x over 1 factorial, plus x squared over 2 factorial plus x cubed over 3 factorial, x to the fourth over 4 factorial, which out to look familiar. Because that is the power series for e to the x. 

And so what we just found, is that the function y, our particular function here, A particular solution to that differential equation is 3e to the x, which is very similar, or it's identical actually, to what we would have found in the previous chapter or two.