# Solve $(x^2 + 1)y'' - 6xy' + 10y =0$ using series method

Use series methods to solve: $(x^2 + 1)y'' - 6xy' + 10y =0$

a) Give the recursion formula

b) Give the first two non-zero terms of the solution corresponding to $a_0 = 1$ and $a_1 = 0$

c) Give the first three non-zero terms of the solution corresponding to $a_0 = 0$ and $a_1 = 1$

Here is what I have so far, then I get stuck on the series when I am trying to get all of my $n = 0$ within each series.

$y = \sum^{\infty}_{n=0} na_nX^n$

$y' = \sum^{\infty}_{n=1} na_nX^{n-1}$

$y'' = \sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}$

so we have: $(x^2+1)[\sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}] - (6x) \sum^{\infty}_{n=1} na_nX^{n-1} + 10\sum^{\infty}_{n=0} na_nX^n$

After doing some series manipulation I got:

$= \sum^{\infty}_{n=2} n(n-1)a_nX^n + \sum^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - \sum^{\infty}_{n=0} 6na_nX^n + \sum^{\infty}_{n=0} 10na_nX^n$

Now I'm stuck on what to do with the first series that has $n=2$ because I don't want to mess up the powers.

• – user64494 Apr 27 '15 at 7:01

I do not know how this is teached but I think that you make the problem difficult changing the starting value of the index in the summation. Let me try $$y=\sum_{n=0}^{\infty}a_n x^n$$ $$y'=\sum_{n=0}^{\infty}n a_n x^{n-1}$$ $$y''=\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}$$ Rewrite the equation as $$(x^2 + 1)y'' - 6xy' + 10y =x^2y''+y''-6xy'+10y=0$$ and include the summations. So, $$\sum_{n=0}^{\infty}n(n-1) a_n x^{n}+\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}-6\sum_{n=0}^{\infty}n a_n x^{n}+10\sum_{n=0}^{\infty}a_n x^n=0$$ Now consider a given power $m$; this gives $$m(m-1)a_m+(m+2)(m+1)a_{m+2}-6ma_m+10a_m=0$$ So, after grouping the terms and simplifying you have $$(m+1)(m+2)a_{m+2}+(m-2)(m-5)a_m=0$$ or $$a_{m+2}=-\frac{(m-2)(m-5)}{(m+1)(m+2)}a_m$$ from which some interesting features appear after a particular value of $m$ (I let you finding it).

• When you take the derivative of $y$ you have to change $n=0$ to $n=1$ because you are taking the derivative, you can't have all of the $n$'s to just stay at $0$, do you see what I'm saying? That is incorrect. Similarly, when you take the second derivative, i.e. $y''$ you have to increase $n$ yet again to $2$ thus: my work above in the OP. – Yusha Apr 27 '15 at 16:42
• So your answer doesn't make any sense, if you could just leave all the summation starting at $0$ then I wouldn't be here in the first place because the problem would be almost trivial, I would just factor and solve for the Recursion Formula. – Yusha Apr 27 '15 at 16:52
• $n a_n=0$ if $n=0$; $n(n-1)a_n=0$ if $n=0$ or $n=1$. So you can let the summation from $n=0$. I am afraid that your statement your answer doesn't make any sense is not totally right. – Claude Leibovici Apr 27 '15 at 17:06
• Yes but when you took the derivative you didn't change the index....How can you just not change the index when doing the derivative? That doesn't make any sense. I understand what you're saying for stripping out terms but you have to take the derivative appropriately right?? – Yusha Apr 27 '15 at 17:08
• I am afraid that you are forgetting your manners. I should prefer you keep your mind open in order to have a positive discussion with you. – Claude Leibovici Apr 27 '15 at 17:10

UPDATE!

The correct solutions for part b) and part c) are:

b)

$a_0 = 1$

$a_1 = 0$

$a_2 = -2/3$

$a_3 = 0$

$a_4 = 0$

$\therefore$ $y_1 = 1 + (-\frac{2}{3}x^2) + ... + ...$

c)

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

$a_3 = -2/3$

$a_4 = 0$

$a_5 = 1/15$

$\therefore$ $y_2 = x + (-\frac{2}{3}x^2) + \frac{1}{15}x^5 + ...$

• Also the recursion formula provided by Claude is correct for part a. – Yusha Apr 28 '15 at 1:59