3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Yusha Apr 27 '15 at 16:42
  • $\begingroup$ 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. $\endgroup$ – Yusha Apr 27 '15 at 16:52
  • 1
    $\begingroup$ $n a_n=0$ 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. $\endgroup$ – Claude Leibovici Apr 27 '15 at 17:06
  • $\begingroup$ 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?? $\endgroup$ – Yusha Apr 27 '15 at 17:08
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Apr 27 '15 at 17:10
0
$\begingroup$

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)

$a_0 = 0$

$a_1 = 1$

$a_2 = 0$

$a_3 = -2/3$

$a_4 = 0$

$a_5 = 1/15$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.