# Closed-form of a power series solution to an ODE

• Feb 22nd 2010, 03:28 AM
chella182
Closed-form of a power series solution to an ODE
Seek a power series solution about $x=0$ of the equation

$(1+x^2)y''+2xy'-2y=0$

as far as the term in $x^6$. Then find a closed-form of the solution (i.e. write the solution using simple functions, in this case arctan, rather than infinite series) by using

$\arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}+...+(-1)^n\frac{x^{2n+1}}{2n+1}+...$.

It's the closed-form bit that's confusing me, 'cause my answer looks nothing like the arctan thing. The answer I got was

$y=A(1+x^2-\frac{1}{3}x^4+\frac{2}{5}x^6+...)$ where A is just a constant.

(Worried)

Can anyone help?
• Feb 22nd 2010, 03:37 AM
CaptainBlack
Quote:

Originally Posted by chella182
Seek a power series solution about $x=0$ of the equation

$(1+x^2)y''+2xy'-2y=0$

as far as the term in $x^6$. Then find a closed-form of the solution (i.e. write the solution using simple functions, in this case arctan, rather than infinite series) by using

$\arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}+...+(-1)^n\frac{x^{2n+1}}{2n+1}+...$.

It's the closed-form bit that's confusing me, 'cause my answer looks nothing like the arctan thing. The answer I got was

$y=A(1+x^2-\frac{1}{3}x^4+\frac{2}{5}x^6+...)$ where A is just a constant.

(Worried)

Can anyone help?

$y=A+Ax(x-\frac{1}{3}x^3+\frac{2}{5}x^5+...)$

CB
• Feb 22nd 2010, 05:05 AM
chella182
Quote:

Originally Posted by CaptainBlack

$y=A+Ax(x-\frac{1}{3}x^3+\frac{2}{5}x^5+...)$

CB

But that still doesn't look like arctan with the $\frac{2}{5}$... unless I'm missing something really obvious here...
• Feb 22nd 2010, 06:25 AM
Opalg
Quote:

Originally Posted by chella182
But that still doesn't look like arctan with the $\frac{2}{5}$... unless I'm missing something really obvious here...

Re-check your answer for the coefficient of $x^5$.
• Feb 22nd 2010, 06:29 AM
chella182
Quote:

Originally Posted by Opalg
Re-check your answer for the coefficient of $x^5$.

I did do, the $\frac{2}{5}$ fits with everything I've done. Obviously wrong then, I'll just leave this question 'cause I'm clearly not gonna get it right now haha.
• Feb 22nd 2010, 06:47 AM
Opalg
Quote:

Originally Posted by chella182
I did do, the $\frac{2}{5}$ fits with everything I've done. Obviously wrong then, I'll just leave this question 'cause I'm clearly not gonna get it right now haha.

I can't comment on where you went wrong because you haven't shown the working. But when I looked for a power series solution $y = \textstyle\sum a_nx^n$ I found that $a_0 = a$ and $a_1=b$ can be arbitrary. Then $a_n=0$ for all odd $n\geqslant3$, $a_2=a$, $a_4 = -\tfrac13a$, and $a_6 = -\tfrac35a_4 = \tfrac15a$.
• Feb 22nd 2010, 07:09 AM
chella182
Yeah well I got wrong last time I showed my working if I remember rightly 'cause this site's not meant for checking work or something, + it would take an age to type it out in latex with all of the sum signs etc.

I got $a_n=0$ for all odd $n$, but then I got $a_0=a_2$ and then $12a_4+4a_2=0$, so $12a_4+4a_0=0$ so rearranging that I got that $a_4=\frac{-a_0}{3}$.

I then got $30a_6+36a_4=0$, and subbing in $a_4=\frac{-a_0}{3}$ I got $30a_6-12a_0=0$, getting that $a_6=\frac{2}{5}a_0$.

(Worried)

Obviously it's just some little mistake somewhere, cheers though. I actually think I can see my mistake... according to me $4\times3=30$ haha! Oh well, sorted now. Thanks again.