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Ok so I am stuck. I need to get all the $n$'s to $=0$ but I can't reduce my series which has $n=2$ to $0$ because then I will have undone all my work in the first place to get all the $X^n$'s to the same power.. I am really confused on what to do.. Here is where I'm at now:

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

any guidance would be great. Thx in advance

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  • $\begingroup$ It is much easier to get all the n's to 2 $\endgroup$ – user174622 Apr 26 '15 at 2:08
  • $\begingroup$ That wouldn't make any sense, because then I wouldnt have my X's to the same power. $\endgroup$ – Yusha Apr 26 '15 at 2:09
  • $\begingroup$ Why would you have to change the powers? $\endgroup$ – user174622 Apr 26 '15 at 2:12
  • $\begingroup$ What do you mean? You can't just change $n =0 $ to $n=2$ without adding $n-2$ to everywhere there is a $n$ right?? $\endgroup$ – Yusha Apr 26 '15 at 2:14
  • $\begingroup$ The powers are all n so it doesn't matter. $\endgroup$ – user174622 Apr 26 '15 at 2:15
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$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + (2a_2+6a_3+\sum ^{\infty}_{n=2} (n+2)(n+1)a_{n+2}X^n) - (a_1+6\sum ^{\infty}_{n=2} na_nX^n) + (a_1+10\sum ^{\infty}_{n=2}a_nX^n)$$

$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_nX^n + (n+2)(n+1)a_{n+2}X^n) - (6 na_nX^n) + (10a_nX^n)]$$

$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_n + (n+2)(n+1)a_{n+2}) - (6 na_n) + (10a_n)]X^n$$

Now you go about solving your differential equation:

$$(2a_2+6a_3)=0\rightarrow a_2=-3a_3$$

$$\sum ^{\infty}_{n=2} [(n^2-7n+10))a_n + (n+2)(n+1)a_{n+2}) ]X^n=0\sum ^{\infty}_{n=2} [(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2}) ]X^n=0$$

$$(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2})=0\rightarrow a_{n+2}=-\frac{(n-5)(n-2)}{(n+2)(n+1)}a_n$$

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  • $\begingroup$ @user3325915 Do you see where I'm going with this? $\endgroup$ – user174622 Apr 26 '15 at 2:23
  • $\begingroup$ Yea I'm looking at it in detail. I see it now. $\endgroup$ – Yusha Apr 26 '15 at 2:27
  • $\begingroup$ @user3325915 You know what to do now right? $\endgroup$ – user174622 Apr 26 '15 at 2:29
  • $\begingroup$ Yea. Give me 4 mins. Working it out $\endgroup$ – Yusha Apr 26 '15 at 2:29
  • $\begingroup$ shouldn;t you have $a_0+a_1 + 10$? $\endgroup$ – Yusha Apr 26 '15 at 2:32

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