Question

Find the series solution (x>0) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also.

Answer 1

\[2xy'' + y' +xy = 0\]

I cant get it to come out correctly. And the textbook does some things with the indecies that are not explained, like not raising the index by 1 when taking the derivative of a series, for example.

Answer 2

I don't follow what they mean by "larger root," but maybe just proceeding with finding the solution will shed some light on that...
\[y=\sum_{n=0}^\infty a_nx^n\\
y'=\sum_{n=1}^\infty na_nx^{n-1}\\
y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\]
I'm assuming series about \(x=0\). Substituting into the equation:
\[2xy''+y'+xy=0\\
2\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_n x^{n+1}=0\]
To match up the indices, I'll pick out the first few terms of the last two series:
\[2\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+\left(a_1+\sum_{n=2}^\infty na_nx^{n-1}\right)+\left(a_0x+a_1x^2+\sum_{n=2}^\infty a_n x^{n+1}\right)=0\]
which can be rewritten as
\[2{\large\sum_{n=2}^\infty} a_nx\left(n^2x^{n-2}+x^{n}\right)+a_1+a_0x+a_1x^2=0\]
Did you find a recursion for the coefficients?

Answer 3

Well, problem is the assumed solution was not even of that form. Normally I would have done exactly as you had, but the textbook said to assume y is of the form:
\[\sum_{n=0}^{\infty}a _{n}x^{r+n}\]Using that as my form, I can come up with various things, but nothing that resembles the correct solution. As for the "roots", you're expected to find roots of a polynomial in r that has been extracted from the sequence.

In an example problem, they have \[2x^{2}y''-xy'+(1+x)y = 0\]The assumed solution form is the one I gave above, and it says y' and y'' are:

\[y' =\sum_{n=0}^{\infty}a_{n}(r+n)x^{r+n-1} \]
\[y''= \sum_{n=0}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n-2}\] I still am unsure why they did not raise the indecies in those derivatives. They plugged in these series forms and came up with:

\[a_{0}[2r(r-1)-r+1]x^{r} + \sum_{n=1}^{\infty}([2(r+n)(r+n-1)-(r+n)+1]a_{n}+a_{n-1})x^{r+n}=0\]

The roots mentioned come from the term outside of the series that uses a0 as its coefficient. If more context is needed, let me know.

Answer 4

" I still am unsure why they did not raise the indecies in those derivatives."

Why?

Answer 5

Usually they would always raise them. In any series solution like problem before they were raised.

Answer 6

Sorry, to be clearer, I think what you're trying to do is what you've done for polynomials of the form x^n. But clearly you can't do that here, and here's why. I'll compare two similar functions, f(x) and g(x) to show you.

\[f(x)=\sum_{n=0}^\infty a_nx^n \\ f'(x)=\sum_{n=0}^\infty na_nx^{n-1}\]
So what happens when we start by plugging in terms to the derivative? that first n is 0, so the whole first term is gone. We can increase the index by one and there's no change here. \[f'(x)=\sum_{n=1}^\infty na_nx^{n-1}\]

Repeat this for a similar one like you are doing: \[g(x)=\sum_{n=0}^\infty a_nx^{n+r} \\ g'(x)=\sum_{n=0}^\infty (n+r)a_nx^{n+r-1}\] Now let's do the same and take out the first term. \[\Large \ g'(x)=ra_0x^{r-1}+\sum_{n=1}^\infty (n+r)a_nx^{n+r-1}\] A different scenario occurs since the whole term is not necessarily zero anymore!

Answer 7

Ah, okay, makes perfect sense, cool. Well, even doing the series using that form, I still am unable to come up with something that works out correctly x_x But good to know that there is no issue with the index now.

Answer 8

Yeah, in these kinds of things there's a ton of ways the algebra can mess up so it's possible lol

Answer 9

Ill give it one more try and see what I can come up with.

Answer 10

I haven't worked the problem yet, I'm scared to currently.

Answer 11

Alright, so I get what I was getting before, so maybe Im just making an error.
\[a_{0}x^{r-1}r(2r-1) + \sum_{n=0}^{\infty}[a_{n+1}(r+n+1)[2(r+n)+1]+a_{n}]x^{r+n}\]

So from the terms removed from the series, I get the polynomial in r to be r(2r-1), giving me roots of 0 and 1/2, which are correct. Now, the recurrence relation that is being set up is wrong from what the book claims. Is there something noticably wrong up to here?

Answer 12

Alright, I came up with the correct recurrence relation and can get an answer:
\[a_{n}=\frac{-a_{n-2}}{(r+n)(3(r+n)-1)}\]So with n starting at 2 and using r = 1/2 (the highest root of the so called "indicial equation"), I am able to get a series in terms of a0 and in terms of a1. Problem now is the book answer omits the series solution that would be in terms of a1, leaving the solution as a series of only even powers. Would anyone have an idea why the odd-powered terms in the series solution would be omitted given the problem or the recurrence relation I have?

This is the only thing left I need explained :P