INSTRUCTOR: Find a power series solution for the following differential equations, part a. The first derivative of y plus 2x times y equals 0. For a power series solution, we're going to begin with a function form, and then solve for some parts of that.

So y of x is equal to the sum from n equals 0 to infinity of a sub n times x to the n. And this means that y prime of x is equal to the series from n equals 1 to infinity of n times a sub n times x to the n minus 1, using the power rule, the general power rule.

Now we're going to substitute these two functions, y of x and y prime of x, into our differential equation. This means that we have the sum from n equals 1 to infinity of n times a sub n times x to the n minus 1 plus 2x times the sum from n equals 0 to infinity of a sub n times x to the n, and that is equal to 0.

First, we'll multiply the coefficient on our second sum into the summand. So this would be the sum from n equals 1 to infinity of n times a sub n times x to the n minus 1, end exponent, plus the sum n equals 0 to infinity of 2 a sub n times x to the n plus 1, end exponent, and that will equal 0.

Now, we want both of these power series that we have to have the same powers of x. So I'm going to adjust my index, index of the second, to be equal to the sum from n equals 1 to infinity of 2 times a sub n minus 1, end subscript, times x to the n. And we'll leave the first power series the same.

Actually, no, we need to adjust that one as well. I'm going to change that to the sum from n equals 0 to infinity of, open parenthesis, n plus 1, close parenthesis, times a sub n plus 1, end subscript, x to the n for the whole thing plus the sum from n equals 1 to infinity of 2 times a sub n minus 1, end subscript, times x to the n equals 0.

Now, because these two series do not begin at the same place, I'm going to remove the 0-th term so that we can combine them. So this would be equal to a sub 1 plus the sum from n equals 1 to infinity of, open bracket, open parenthesis, n plus 1, close parenthesis, a subscript of n plus 1, end subscript, plus 2 a subscript of n minus 1, end subscript, close bracket, x to the n equals 0.

Now, we can equate coefficients to come up with a relationship for these. So because this is equal to-- identically equal to 0, a sub 1 must be equal to 0. I'll just add that here. a sub 1 is equal to 0. But that also means that n plus 1-- or parenthesis, n plus 1, close parenthesis multiplied by a sub n plus 1, end subscript, plus 2 a sub n minus 1, end subscript, is equal to 0, as long as n is greater than or equal to 1, based on our index.

Now, that's actually a recursive relationship. So we can start evaluating for specific values of n to find out what some of our coefficients are. So that should-- but we need to hold on to that a sub 1 is equal to 0. Now let me just rearrange this recursive relationship to say that a sub n plus 1, end subscript, is equal to 2-- or negative 2 times a n minus 1, end subscript, over the quantity n plus 1.

So if I take n equal to 1, for instance, n equal to 1 would tell me that a sub 2 is equal to negative 2 a sub 0 all over 2, should be equal to negative a0. OK. So a sub 2 is equal to negative a sub 0. OK, that's fine.

a sub 3 is then equal to, well, negative 2 times a sub 1 over-- well, it doesn't really matter because it's going to be equal to 0, because a sub 1 is equal to 0. So a sub 3 is equal to 0. And then, if you notice, because these two-- these terms are tied with the relationship of 2-- so n minus 1 as opposed n plus 1-- it actually turns out that all of the odd terms will be equal to 0, because a5 will be related to a3, which is equal to 0. a7 will be related to a5, which is related to a3, which will be equal to 0.

So we find this tracing back all the way. So this is actually true for all of the odd terms. So let's concentrate just a little bit on a sub 4. Let me just actually write these down here. So a sub 1 is equal to 0. a sub 2 is equal to negative 2-- or, sorry, negative a sub 0. a sub 0 was actually unknown. That's actually going to be our constant in this whole matter, this differential equation.

a sub 4 will be equal to negative 2 times a sub 2, which is negative a sub 0. And that's all over 4. So that is equal to a sub 0-- or 1/2 a sub 0. Now, a sub 6 will be related to that. So a sub 6 will be equal to negative 2 times a sub 4, which is actually equal to 1/2 a sub 0. And that will be over 6. So, simplifying that, we see that a sub 6 is equal to negative 1 over-- and I'm going to write that as 3 times 2, end fraction, times a sub 0.

All right, a sub 8-- at this point, we may notice a pattern and actually write out the function. a sub 8 is equal to negative 2 multiplied by negative 1 over 3 times 2, end fraction, a sub 0. And all of that is over 8. Which becomes-- all right. So that becomes 1 over 4 times 3 times 2, end fraction, times a sub 0.

So, as I said, we may notice a pattern at this point. And I'm going to go ahead and write out that y of x is equal to a sub 0, because that's unknown, minus a sub 0 times x squared plus 1 over 2, 1/2, times a sub 0 x to the fourth minus 1 over 3 times 2, end fraction, a sub 0 x to the sixth plus 1 over 4 times 3 times 2, end fraction, a sub 0 x to the eighth. And it continues in that pattern.

Now, since every term has an a sub 0, this is going to be equal to a sub 0 multiplied by the sum from n equals 0 to infinity of the fraction, parenthesis, negative 1, close parenthesis, to the n, denominator of n factorial, end fraction, x to the 2n, end exponent. Now pause this and take a moment to convince yourself that actually does hold with the pattern we've already observed.

And now that you've done that, this is actually a known power series. We know what e to the x is. Well, here, this is actually e to the x with an inserted negative x squared. So this will be equal to a sub 0 times e to the exponent of negative x squared, end exponent. So that, finally, our solution, our power series solution, is y of x equals a sub 0 times e to the exponent of negative x squared, end exponent. Or, if you would rather, the power series that we had just a moment ago.