INSTRUCTOR: Find the power series for this differential equation, part b. Open parenthesis, x plus 1, close parenthesis, multiplied by the first derivative of y equals 3 times y. Now, again, we will go back and say y of x is equal to the sum from n equals 0 to infinity of a sub n times x to the n. And y of x-- or y prime of x is equal to the sum from n equals 1 to infinity of n times a sub n times x to the exponent of n minus 1, end exponent.

Now let's actually replace this function-- it's a generic function because we don't know the coefficients yet. Let's replace that into our differential equation. So that we have, open parenthesis, x plus 1, close parenthesis, multiplied by the series from n equals 1 to infinity of n times a sub n times x to the exponent of n minus 1, end exponent. And that will be equal to 3 times the series from n equals 0 to infinity of a sub n times x to the n.

Let's go ahead and multiply our coefficients in. This actually will become a couple of-- we're going to expand the x plus 1, the binomial term, into a series, so I'm going to end up having three terms. This will be the sum from n equals 1 to infinity of n times a sub n times x to the n plus the series or the sum from n equals 0 to infinity-- or, actually, sorry. That's n equals 1 to infinity of n times a sub n times x to the n minus 1. And that will be equal to the sum from n equals 0 to infinity of 3 a sub n times x to the n.

Now let's adjust our indices so that we have all powers of x to the n. Also, we only need to mess with the second term. So this second series will be equal to the sum from n equals 0 to infinity of, open parenthesis, n plus 1, close parenthesis, a sub n plus 1, end subscript, times x to the n. And we'll bring down the other two power series.

Now, at this point, we need to remove the 0-th terms so that our starting values are the same. And, in fact, that would be the-- yes, we'll do that with the second and third power series in our equation. So this will be equal to a sub 1 plus the series n equals 1 to infinity, bracket, n times a sub n plus, open parenthesis n plus 1, close parenthesis, a sub n plus 1, end subscript, end bracket, x to the n equals 3 a sub 0 plus the sum from n equals 1 to infinity of 3 times a sub n times x to the n.

Now, we have two sides in the equation, so we want to equate coefficients here. First thing we notice is that a sub 1 equals 3 a sub 0. And then, also, n times a sub n plus, open parenthesis, n plus 1, close parenthesis, multiplied by a sub n plus 1, end subscript, is equal to 3 a sub 0, as long as n is greater than or equal to 1. And these coefficients will hold in that recursive relationship.

Now, that means that a sub n plus 1, end subscript, is equal to a fraction-- open parenthesis, 3 minus n, close parenthesis, times a sub n. That was our numerator. With a denominator of n plus 1, end fraction.

So let's start using some values of n. Since this relationship is valid for n greater than or equal to 1, let's take the value of 1. If n is equal to 1, then this means that a sub 2 is equal to 2 times a sub 1 over 2. And since a sub 1 is equal to 3 a sub 0, that is equal to 3 a sub 0.

How about a3? That would be an n value of 2. a3 is equal to a sub 2 over 3, which, coincidentally, is equal to a sub 0. And then, let's see. a sub 4 actually is equal to 0. And because future terms depend on the previous term, all the rest of our terms will be equal to 0.

So we can write y of x is equal to-- so let me put it up here. y of x is equal to a sub 0 plus 3 a sub 0 x plus 3 a sub 0 x squared plus a sub 0 x cubed. And they all have an a sub 0 term in common. And then we can also actually factor this. It's going to be a, well, sum of cubes. Well, no, that's not quite the right term.

But we can factor this, and so it becomes y of x is equal to a sub 0 times, open parenthesis, 1 plus x, close parenthesis, cubed. It's based off of a binomial series. Well, anyways, y of x, our general solution, is equal to a sub 0, parenthesis, 1 plus x, close parenthesis, cubed. And that is a power series solution. It happens to be a finite function, finitely termed function, but it is a power series solution to that differential equation.