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Q: By seeking a power series solution to $$2xy′′+(3−x)y′−y = 0$$ about $x=0$ show that there are two linearly independent solutions that have the recurrence relations $$a_{n+1} =\frac{a_n}{2n+3}$$ and $$b_{n+1} =\frac{b_n}{2n+2}$$ governing the coefficients of the terms in their respective series.


Attempt:

$y=\sum\limits_{m=0}^\infty {a_mx}^{m+r}$

$y'=\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}$

$y''=\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}$


Substitute:

$2x\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2} +(3-x)\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1} -\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$


Reduce:

$\sum\limits_{m=0}^\infty {2(m+r-1)}{(m+r)}{a_mx}^{m+r-1} +\sum\limits_{m=0}^\infty {3(m+r)}{a_mx}^{m+r-1} -\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r} -\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$


Combine:

$\sum\limits_{m=0}^\infty[2(m+r-1)(m+r)+3(m+r)]a_{m}x^{m+r-1} -\sum\limits_{m=0}^\infty [(m+r)+1]{a_mx}^{m+r}=0$


Shift Index:

This is as far as I got, the method I was taught is this:

Let $$m'= m-1$$ Then $$m=m'+1$$ In the summation above every where there is an $m$ substitute $m'$ rearrange the formula and drop the prime.


My confusion is this, if I do this shift with the first summation then I get

$\sum\limits_{m=-1}^\infty[2(m+r)(m+r+1)+3(m+r+1)]a_{m+1}x^{m+r}$

Then from this how do I get the indicial equation and then how do I use that to gain the recurrence relation? I don't have to build the power series in this part of the question, but that method I can do, you use the recurrence relation, substitute your $r$ values and get the first three terms, and collect to build the power series with respect to two independent terms which are usually $a_0$ & $a_1$.

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  • $\begingroup$ Have you checked if you have irregular singular points, that might be the problem if the summation isn't working out. $\endgroup$ – HELP May 15 '16 at 16:34
  • $\begingroup$ Since $m' - m \in \mathbb Z^+$ consider Fuchs theorem to obtain the two independent solutions. Then calculate $W(y_1,y_2 | x)$ to assure that it is nonzero $\forall x \in I$. Another method is to obtain $y_1$ and use order reduction to obtain $y_2$ $\endgroup$ – Rebellos May 15 '16 at 16:34
  • $\begingroup$ @CharalamposFilippatos this is definitely 2nd year maths, i think Fuchs theorem is a bit too advanced $\endgroup$ – HELP May 15 '16 at 16:40
  • $\begingroup$ Yes this is second year DE, I know how to prove independence with the wronskian, but I don't know the Fuchs theorem. $\endgroup$ – UniStuffz May 15 '16 at 16:41
  • $\begingroup$ I am a first year student at NTUA and we have studied such differential equations. Fuchs theorem en.wikipedia.org/wiki/Fuchs%27_theorem $\endgroup$ – Rebellos May 15 '16 at 16:41
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I've never seen an shifting where you go from a negative to a positive. The equation you gain for your r values are given by the lowest power of x. So if you were to shift from -1, then remove your n=-1 term, gain the equation to find r, then combine your summations all starting from zero. That should give you a general x^n+r term, which you set the coefficients to zero, then solve for the recurrence relation. having said that i have no idea how to separate the coefficients into two recurrence relationships.

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