Frobenius’ Method to Solve an ODE (Hydrogen Atom - Radial Equation) My goal is to find two linearly independent solutions to the ODE
$$
r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}+[r^2+\lambda r-l(l+1)]R=0
$$
in the interval $[0,\infty)\ni r$, where $R=R(r)$, and $\lambda\in\mathbb{R}_+$ and $l\in\mathbb{Z}_0$ are fixed constants. Just to put in context my problem, this ODE determined the radial component of the wavefunction of an electron in an atom. The standard approach (e.g. here) is to investigate the asymptotic behaviour of the equation for large and small $r$, factor out this behaviour and then use a series to try to find the solution of what's left. Nonetheless, I'm looking for an answer of the form: ''The general solution to this ODE is given by $R(r)=c_1y_1(r)+c_2y_2$, but since $R(0)$ must be well-defined and $y_2(r)\to-\infty$ as $r\to0$, then $c_2=0$ for the answer to be physically acceptable''. I know that the ''physically acceptable'' solution is given by
$$
y_1\propto\left(\frac{2r}{n}\right)^le^{-r/n}L^{2l+1}_{n+l}\left(\frac{2r}{n}\right)
$$
for all $n\in\mathbb{Z}_+$, and where $L^{2l+1}_{n+l}$ are the associated Laguerre polynomials.
My attempt to find $y_1$ and $y_2$ is to use Frobenius’ method, but i keep stuck at some point: since $r=0$ is a regular singular point of the differential equation, there exists at least one solution of the form
$$
R(r)=\sum_{k=0}^\infty c_kr^{k+s}
$$
where $s$ are indicial roots to be found. Then
$$
\frac{dR}{dr}(r)=\sum_{k=0}^\infty (k+s)c_kr^{k+s-1}\quad\text{and}\quad\frac{d^2R}{dr^2}(r)=\sum_{k=0}^\infty (k+s)(k+s-1)c_kr^{k+s-2}
$$
and so, substituting in the equation, it yields
\begin{align} 
0&=r^s\Biggl[c_0[s(s+1)-l(l+1)]+(c_1[(s+1)(s+2)-l(l+1)]+\lambda c_0)r \\
 &\quad+\sum_{k=2}^\infty\Bigl(c_k[(k+s)(k+s-1)+2(k+s)-l(l+1)]+c_{k-2}+\lambda c_{k-1}\Bigr)r^k\Biggr] 
\end{align}
The first term gives me the indicial roots
$$
c_0[s(s+1)-l(l+1)]=0\quad\Rightarrow\quad s_1=l\quad\text{and}\quad s_2=-l-1
$$
and because nothing is gained by taking $c_0=0$. Now, since $s_1$ and $s_2$ are distinct and the difference $s_1-s_2=2l+1$ is a positive integer, then there exist two linearly independent solutions of the form
$$
R^1(r)=\sum_{k=0}^\infty c_kr^{k+l},\quad c_0\neq0
$$
and
$$
R^2(r)=CR^1(r)\ln(r)+\sum_{k=0}^\infty b_kr^{k-l-1}
$$
where $C$ is a constant that could be zero (I'm following the book ''Differential Equations with Boundary-Value Problems; DENNIS G. ZILL and MICHAEL R. CULLEN''.) For the second term
$$
c_1[(s+1)(s+2)-l(l+1)]+\lambda c_0=0
$$
from where 
$$
c_1=\frac{l+l^2-\lambda c_0}{2+3l+l^2}\quad\text{for $s_1=l$},\quad\text{and}\quad c_1=\frac{l+l^2-\lambda c_0}{l(l-1)}\quad\text{for $s_2=-l-1$}
$$
For $k\geq2$, I have
$$
c_k[(k+s)(k+s-1)+2(k+s)-l(l+1)]+c_{k-2}+\lambda c_{k-1}=0
$$

How can I continue from here to find the solution/series $R^1(r)$ and $R^2(r)$ ? How are they ?
Is there a better way to proceed in order to find $R^1(r)$ and $R^2(r)$ ?

Thanks in advance.
 A: Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=250:
Let $R=r^kS$ ,
Then $\dfrac{dR}{dr}=r^k\dfrac{dS}{dr}+kr^{k-1}S$
$\dfrac{d^2R}{dr^2}=r^k\dfrac{d^2S}{dr^2}+kr^{k-1}\dfrac{dS}{dr}+kr^{k-1}\dfrac{dS}{dr}+k(k-1)r^{k-2}S=r^k\dfrac{d^2S}{dr^2}+2kr^{k-1}\dfrac{dS}{dr}+k(k-1)r^{k-2}S$
$\therefore r^2\left(r^k\dfrac{d^2S}{dr^2}+2kr^{k-1}\dfrac{dS}{dr}+k(k-1)r^{k-2}S\right)+2r\left(r^k\dfrac{dS}{dr}+kr^{k-1}S\right)+(r^2+\lambda r-l(l+1))r^kS=0$
$r^{k+2}\dfrac{d^2S}{dr^2}+2kr^{k+1}\dfrac{dS}{dr}+k(k-1)r^kS+2r^{k+1}\dfrac{dS}{dr}+2kr^kS+(r^2+\lambda r-l(l+1))r^kS=0$
$r^{k+2}\dfrac{d^2S}{dr^2}+2(k+1)r^{k+1}\dfrac{dS}{dr}+(r^2+\lambda r+k(k+1)-l(l+1))r^kS=0$
$r^2\dfrac{d^2S}{dr^2}+2(k+1)r\dfrac{dS}{dr}+(r^2+\lambda r+k(k+1)-l(l+1))S=0$
Choose $k=l$ , the ODE becomes
$r^2\dfrac{d^2S}{dr^2}+2(l+1)r\dfrac{dS}{dr}+(r^2+\lambda r)S=0$
$r\dfrac{d^2S}{dr^2}+2(l+1)\dfrac{dS}{dr}+(r+\lambda)S=0$
