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This is actually a very simple problem, but I am going to type everything out in case I really overlooked something

I am trying to find a recurrence relation in the series solution I got. The ODE is $y'' + xy'- y = 0$.

$\begin{align} y'' + xy' - y & = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} a_nx^n \\ & = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + \sum_{n=1}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n \\ & = \sum_{n=0}^{\infty} (n+2)(n+1)a_nx^n + \sum_{n=0}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n\\ & = \sum_{n=0}^{\infty} x^n[(n+1)(n+2)a_{n+2} - a_n + na_n ] \\ & = 0 \end{align}$

The power series is 0 iff $(n+1)(n+2)a_{n+2} - a_n(1- n) = 0 \iff a_{n+2} = \dfrac{a_n (1 - n)}{(n+1)(n+2)}$

So testing out various values of n = 1, 2, 3... led me to

$a_2 = \dfrac{a_0}{2}$

$a_3 = 0$

$a_4 = \dfrac{-a_2}{12} = \dfrac{a_0}{24}$

$a_5 = 0$

$a_6 = -\dfrac{a_4}{10} = \dfrac{a_0}{240}$

Okay so clearly the odd terms all go to 0 and there will be no odd terms. I know that the even terms alternate back and forth in series (so I will get an alternating series). I have no idea how the denominator can jump from 2 to 24 and then to 240 and then to a stunning 240*56. I see no pattern in how the denominator jumps at all, it jumps too fast if you ask me

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    What's the ODE here? is it $y''+xy'+y=0$ or $y''+xy'-y=0$?2012-06-21
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    @Mercy, sorry it is $y'' +xy' - y= 0$2012-06-21
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    Don't be too hasty with "there will be no odd terms": $a_1$ is arbitrary (and in fact $y=x$ is a solution). Remember that for a second-order differential equation there must be two arbitrary constants in the general solution.2012-06-21
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    Oh wow, I had completely forgotten that $a_1$ existed2012-06-21

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Rewriting the recurrence $$a_{n+2}=\frac{a_n(1-n)}{(n+1)(n+2)}=-\frac{a_n(n-1)}{(n+1)(n+2)}$$ as $$a_{n+2}(n+1)=-\frac{a_n(n-1)}{n+2}$$ suggests the substitution $b_n=a_n(n-1)$, $b_0=-a_0$. Then we have simply $$b_{n+2}=-\dfrac{b_n}{n+2}\;.$$

Sticking to the even-numbered terms, we get $b_0,b_2=-\dfrac{b_0}2,b_4=\dfrac{b_0}{2\cdot4},b_6=-\dfrac{b_0}{2\cdot4\cdot6}$, and it’s apparent (and easily proved by induction) that $$b_{2n}=\frac{(-1)^nb_0}{2^nn!}\;.$$

This translates back to $$a_{2n}=\frac{(-1)^{n+1}a_0}{(2n-1)2^nn!}\;.$$

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    The second equation should be $$a_{n+2}(n+1)=-\dfrac{a_n(n-1)}{n+2}. $$2012-06-21
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    @Stefanos: Thanks!2012-06-21
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    @BrianM.Scott, what is motivation behind writing it as $a_{n+2}(n+1)=-\frac{a_n(n-1)}{n+2}$? I was taught to always solve for the highest subscript2012-06-21
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    @jak: I was looking for some way to simplify it and noticed that if I pulled the $n+1$ over to the other side, I had (bar the minus sign) something of the form $a_k(k-1)$ on each side. That meant that I could substitute and get a much simpler recurrence. If you’ve never seen the trick of substituting to get a simpler series, it probably isn’t at all obvious.2012-06-21
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    @BrianM.Scott, to find the new $b_{n+2}$, is this what you did? $b_n = a_n(n-1) \implies b_{n+2} = a_{n+2}(n + 2 - 1) = a_{n+2}(n + 1)$2012-06-21
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    @jak: If $b_n=a_n(n-1)$, then $b_{n+2}=a_{n+2}(n+1)$, so $a_{n+2}(n+1)=-\frac{a_n(n-1)}{n+2}$ becomes $b_{n+2}=-\dfrac{b_n}{n+2}$ by direct substitution.2012-06-21
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    @jak: No, $b_n=a_n(n-1)$ implies that $b_{n+2}=a_{n+2}(n+2-1)$.2012-06-21
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    @BrianM.Scott, isn't that exactly what I did?2012-06-21
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    @jak: It is now, yes; but I think that before you edited your comment, you had $b_{n+1}$ instead of $b_{n+2}$. (If not, my apologies for misreading.)2012-06-21
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    Ah yes I did Brain, my usual typos. Also, since $a_0$ and $a_1$ is free in the problem, should I set it to 1 (was going to set it to 0, but that would give me the trivial solution)2012-06-21
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    @jak: Any value of $a_0$ will give you a legitimate solution, so unless there are boundary conditions of some kind, you should just leave it as is, rather like a constant of integration.2012-06-21
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    Just to make sure I really wrote this correctly and didn't make any careless mistake (as you've seen already...). I got my sum to be $y = a_1 + a_0\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^n}{(2n-1)2^nn!}$2012-06-21
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    @jak: Looks okay to me. (Mind you, I’ve been known to make careless mistakes myself.)2012-06-21
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A start: Everything will go more nicely if we avoid premature cancellation.

Let's forget about $a_0$, it is a constant multiplier, we can put it in later. (If I were doing this for myself, I would also forget temporarily about minus signs, they are under control. And $1-n$ really wants to be written as $-(n-1)$.)

When we calculate without cancellation we get $a_2=-\frac{1}{1\cdot 2}$, $\,a_4=\frac{1}{1\cdot 2\cdot 3\cdot 4}$, $\,a_6=-\frac{1\cdot 3}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6}$. The next term has $1\cdot 3\cdot 5$ on top.

That should be roughly enough to see what's going on. If not, one or two more terms are decisive.

We get the obvious factorials at the bottom. At the top, we have the product of successive odd numbers. It may be useful to simplify this. For example, look at $1\cdot 3\cdot 5\cdot 7$, which comes up a little later. This is $8!$ with the even terms missing. So it is $\frac{8!}{2\cdot 4\cdot 6\cdot 8}$, which is $\frac{8!}{(2^4)(4!)}$.

Once we know the answer, we can, if we are in the mood, provide a formal proof by induction.

Remarks: $1.$ Products such as $7\cdot 5\cdot 3\cdot 7$ or $10\cdot 8\cdot 6\cdot 4\cdot 2$ come up often enough that a special notation has been devised for them. the first is denoted by $7!!$, the second by $10!!$. This is a somewhat strange notation, since $7!!$ is definitely not the same as $(7!)!$.

$2.$ Premature cancellation can hide structure. There is an associated more modern affliction called digititis. Everything gets pushed into a calculator. The result is usually a structureless jumble of digits.

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It is obvious that $a_{2n+1}=0$ for every $n \ge 1$. Set $b_n=a_{2n}$ for $n \ge 0$. Then $$ b_k=a_{2(k-1)+2}=-\frac{2k-3}{(2k-1)(2k)}b_{k-1}=-\frac{1}{2}\cdot\frac{1}{k}\cdot\frac{2k-3}{2k-1}b_{k-1} \quad \forall k \ge 0. $$ Taking the product on both sides for $1 \le k \le n$ we get $$ b_nb_{n-1}\cdots b_1=\left(-\frac{1}{2}\right)^{n}\cdot\frac{1}{n!}\cdot\frac{-1.3.\cdots(2n-3)}{1.3.\cdots(2n-3)(2n-1)}b_{n-1}b_{n-2}\cdots b_1b_0. $$ After simplication we get $$ b_n=\left(-\frac{1}{2}\right)^{n}\cdot\frac{-1}{n!(2n-1)}b_0, $$ i.e. $$ a_{2n}=\frac{(-1)^{n+1}}{n!2^n(2n-1)}a_0. $$

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    how did you change the subscript in the first line there? I am a little confused.2012-06-21
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    Do you mind explaining again how you got from $b_n = a_{2n}$ to $b_k = a_{2(k-1)+2}$?2012-06-22
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    $2(k-1)+2=2k-2+2=2k$2012-06-22