# Frobenius Method recurrence relations

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 $a_0$ & $a_1$.

• Have you checked if you have irregular singular points, that might be the problem if the summation isn't working out. – HELP May 15 '16 at 16:34
• 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 $y_2$ – Rebellos May 15 '16 at 16:34
• @CharalamposFilippatos this is definitely 2nd year maths, i think Fuchs theorem is a bit too advanced – HELP May 15 '16 at 16:40
• Yes this is second year DE, I know how to prove independence with the wronskian, but I don't know the Fuchs theorem. – UniStuffz May 15 '16 at 16:41
• I am a first year student at NTUA and we have studied such differential equations. Fuchs theorem en.wikipedia.org/wiki/Fuchs%27_theorem – Rebellos May 15 '16 at 16:41