<s>[INST] Show that $r^l, r^{-l-1}$ are solutions to $\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=l(l+1)$ Here is a standard textbook question that is causing great difficulty (even though the question says it is easy):

I'm only interested in finding the solution to the second of the above $2$ differential equations by using a power series method:
$$\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=l(l+1)\tag{1}$$
which has solutions $$R=\begin{cases} r^l \\  r^{-l-1} \end{cases}\tag{2}$$ (as printed in the back of the book)
Attempt #1:
Let $$R=\sum_{n=0}^\infty C_n\cdot r^n\tag{3}$$ where the $C_n$ are the expansion coefficients.
Then $$\frac{dR}{dr}=\sum_{n=1}^\infty C_n\cdot n\,r^{n-1}\tag{4}$$
and $$\frac{d^2R}{dr^2}=\sum_{n=2}^\infty C_n \cdot n(n-1)r^{n-2}\tag{5}$$
Re-writing $(1)$ via the product rule
$$2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}=R\,l(l+1)\tag{6}$$
Now substituting $(3)$, $(4)$ and $(5)$ into $(6)$
$$2r\sum_{n=1}^\infty C_n \cdot n\,r^{n-1}+r^2\sum_{n=2}^\infty C_n \cdot n(n-1)\,r^{n-2}=l(l+1)\sum_{n=0}^\infty C_n \cdot r^{n}\tag{7}$$
I am unsure how to proceed from here to show that $$R=\begin{cases} r^l \\  r^{-l-1} \end{cases}\tag{2}$$
Could someone please give me some hints or advice on how I can proceed to reach $(2)$?

EDIT:
Proceeding with the advice given in the first comment by using the method of Frobenius:
Attempt #2:
Letting $$R=\sum_{n=0}^\infty C_n\cdot r^{s+n}\tag{A}$$ where the $C_n$ are the expansion coefficients. 
Then $$\frac{dR}{dr}=\sum_{n=0}^\infty C_n\cdot (s+n)\,r^{s+n-1}\tag{B}$$
and $$\frac{d^2R}{dr^2}=\sum_{n=0}^\infty C_n \cdot (s+n)(s+n-1)\,r^{s+n-2}\tag{C}$$
Writing out the sums explicitly:
$$R=C_0\cdot r^s+C_1\cdot r^{s+1}+C_2\cdot r^{s+2}+C_3\cdot r^{s+3}+\cdots+C_n\cdot r^{s+n}$$
$$\frac{dR}{dr} = C_0\cdot s\,r^{s-1}+C_1\cdot(s+1)\,r^s + C_2\cdot(s+2)\,r^{s+1}+C_3\cdot(s+3)\,r^{s+2}\quad+\cdots + C_n\cdot(s+n)\, r^{s+n-1}$$
$$\frac{d^2R}{dr^2} = C_0\cdot s(s-1)\,r^{s-2}+C_1\cdot s(s+1)\,r^{s-1} + C_2\cdot(s+1)(s+2)\,r^s\quad +C_3\cdot(s+2)(s+3)\,r^{s+1}+\cdots + C_n\cdot(s+n)(s+n-1)r^{s+n-2}$$
Therefore
$$2r\frac{dR}{dr}=2C_0\cdot sr^s+2C_1\cdot (s+1)r^{s+1}+2C_2\cdot(s+2) r^{s+2}+2C_3\cdot(s+3)r^{s+3}\quad+\cdots+2C_n\cdot(s+n) r^{s+n}$$ 
$$r^2\frac{d^2R}{dr^2} = C_0\cdot s(s-1)\,r^{s}+C_1\cdot s(s+1)\,r^{s+1} + C_2\cdot(s+1)(s+2)\,r^{s+2}\quad + C_3\cdot(s+2)(s+3)\,r^{s+3}+\cdots + C_n\cdot(s+n)(s+n-1)\, r^{s+n}$$
$$l(l+1)R=l(l+1)C_0\cdot r^s+l(l+1)C_1\cdot r^{s+1}+l(l+1)C_2\cdot r^{s+2}+l(l+1)C_3\cdot r^{s+3}+\cdots+l(l+1)C_n\cdot r^{s+n}$$
Now rewriting equation $(6)$:
$$2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}=R\,l(l+1)$$
in terms of summations gives
$$2r\sum_{n=0}^\infty C_n \cdot (s+n)\,r^{s+n-1}+r^2\sum_{n=0}^\infty C_n \cdot (n+s)(n+s-1)\,r^{s+n-2}=l(l+1)\sum_{n=0}^\infty C_n \cdot r^{s+n}$$
by bringing in the $r$'s into the sums for the first two summations on the LHS I find that
$$2\sum_{n=-1}^\infty C_{n+1}  (s+n+1)r^{s+n}+\sum_{n=-2}^\infty C_{n+2}(s+n+2)(s+n+1)r^{s+n}=l(l+1)\sum_{n=0}^\infty C_n  r^{s+n}$$
by shifting the index from $n\rightarrow n-1$ and $n\rightarrow n-2$ for the first and second sums respectively.
Now that the powers of $r$ are the same I can compare coefficients:
For $$n \ge -1\implies 2C_{n+1}\cdot(s+n+1)=l(l+1)\cdot C_n$$
For $$n \ge -2\implies 2C_{n+1}\cdot(s+n+1)+C_{n+2}\cdot(s+n+2)(s+n+1)=l(l+1)\cdot C_n$$
But how do I get the Indical equation from here to find the value of $s$?

Thank you.
 [/INST] Did has already explained in the comments that it makes more sense to use a single power ansatz than a power series. If you really want to use a power series, you have to include negative powers if you want to get the $r^{-l-1}$ solution, since this can't be written as a series of non-negative powers at $r=0$.
In the equation you derived, pulling the extra powers of $r$ into the sums leads to all three sums having the same power $r^n$. Then equating the coefficients for all powers yields
$$
C_n(2n+n(n-1))=C_nl(l+1)\;,
$$
or
$$
C_n(n(n+1)-l(l+1))=0\;.
$$
It follows that for each $n$ either $C_n=0$ or $n(n+1)=l(l+1)$. The latter equation has the two solutions $n=l$ and $n=-l-1$, so those are the only two powers that can have non-zero coefficients, and the general solution is a linear combination of the two.
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