The general solution for hydrogen
In Eq. (19.35) we have written the wave functions for the hydrogen atom as \begin{equation} \label{Eq:III:19:37} \psi_{l,m}(\FLPr)=a\,Y_{l,m}(\theta,\phi)F_l(r). \end{equation} These wave functions must be solutions of the differential equation (19.7). Let’s see what that means. Put (19.37) into (19.7); you get  \begin{align} \frac{Y_{l,m}}{r}\,\frac{\partial^2}{\partial r^2}\, (rF_l)+\frac{F_l}{r^2\sin\theta}\,\ddp{}{\theta}&\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{F_l}{r^2\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}\notag\\[1ex] \label{Eq:III:19:38} &=\;-\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr) Y_{l,m}F_l. \end{align} 
 \begin{align} \frac{Y_{l,m}}{r}\frac{\partial^2}{\partial r^2} (rF_l)\!&+\!\frac{F_l}{r^2\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\notag\\[1ex] \label{Eq:III:19:38} &+\!\frac{F_l}{r^2\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}\\[1.5ex] &=-\frac{2m}{\hbar^2}\biggl(\!E+\!\frac{e^2}{r}\!\biggr) Y_{l,m}F_l.\notag \end{align}  Now multiply through by $r^2/F_l$ and rearrange terms. The result is  \begin{align} \frac{1}{\sin\theta}\,\ddp{}{\theta}&\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{1}{\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}\notag\\[1ex] \label{Eq:III:19:39} &=\;-\biggl[ \frac{r^2}{F_l}\biggl\{ \frac{1}{r}\,\frac{d^2}{dr^2}\,(rF_l)+\frac{2m}{\hbar^2} \biggl(E+\frac{e^2}{r}\biggr)F_l\biggr\}\biggr]Y_{l,m}. \end{align} 
 \begin{equation} \begin{gathered} \frac{1}{\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\!+\! \frac{1}{\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}\\[2ex] =-\biggl[ \frac{r^2}{F_l}\biggl\{\! \frac{1}{r}\frac{d^2}{dr^2}(rF_l)\!+\!\frac{2m}{\hbar^2} \biggl(\!E+\frac{e^2}{r}\!\biggr)F_l\!\biggr\}\biggr]Y_{l,m}. \end{gathered} \label{Eq:III:19:39} \end{equation}  The left-hand side of this equation depends on $\theta$ and $\phi$, but not on $r$. No matter what value we choose for $r$, the left side doesn’t change. This must also be true for the right-hand side. Although the quantity in the square brackets has $r$’s all over the place, the whole quantity cannot depend on $r$, otherwise we wouldn’t have an equation good for all $r$. As you can see, the bracket also does not depend on $\theta$ or $\phi$. It must be some constant. Its value may well depend on the $l$-value of the state we are studying, since the function $F_l$ must be the one appropriate to that state; we’ll call the constant $K_l$. Equation (19.39) is therefore equivalent to two equations:  \begin{gather} \label{Eq:III:19:40} \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{1}{\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}=-K_lY_{l,m},\\ \notag\\[2pt] \label{Eq:III:19:41} \frac{1}{r}\,\frac{d^2}{dr^2}\,(rF_l)+ \frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)F_l= K_l\,\frac{F_l}{r^2}. \end{gather} 
  \begin{align} \frac{1}{\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\!+\! \frac{1}{\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}&\notag\\[1.5ex] \label{Eq:III:19:40} =-K_lY_{l,m},&\\[2.5ex] \frac{1}{r}\frac{d^2}{dr^2}(rF_l)+ \frac{2m}{\hbar^2}\biggl(\!E+\!\frac{e^2}{r}\!\biggr)F_l&\notag\\[.5ex] \label{Eq:III:19:41} =K_l\frac{F_l}{r^2}.& \end{align}  
 Now look at what we’ve done. For any state described by $l$ and $m$, we know the functions $Y_{l,m}$; we can use Eq. (19.40) to determine the constant $K_l$. Putting $K_l$ into Eq. (19.41) we have a differential equation for the function $F_l(r)$. If we can solve that equation for $F_l(r)$, we have all of the pieces to put into (19.37) to give $\psi(\FLPr)$.  What is $K_l$? First, notice that it must be the same for all $m$ (which go with a particular $l$), so we can pick any $m$ we want for $Y_{l,m}$ and plug it into (19.40) to solve for $K_l$. Perhaps the easiest one to use is $Y_{l,l}$. From Eq. (18.24), \begin{equation} \label{Eq:III:19:42} R_z(\phi)\,\ket{l,l}=e^{il\phi}\,\ket{l,l}. \end{equation} The matrix element for $R_y(\theta)$ is also quite simple: \begin{equation} \label{Eq:III:19:43} \bracket{l,0}{R_y(\theta)}{l,l}=b\,(\sin\theta)^l, \end{equation} where $b$ is some number.3  Combining the two, we obtain \begin{equation} \label{Eq:III:19:44} Y_{l,l}\propto e^{il\phi}\sin^l\theta. \end{equation} Putting this function into (19.40) gives \begin{equation} \label{Eq:III:19:45} K_l=l(l+1). \end{equation}  Now that we have determined $K_l$, Eq. (19.41) tells us about the radial function $F_l(r)$. It is, of course, just the Schrödinger equation with the angular part replaced by its equivalent $K_lF_l/r^2$. Let’s rewrite (19.41) in the form we had in Eq. (19.8), as follows: \begin{equation} \label{Eq:III:19:46} \frac{1}{r}\frac{d^2}{dr^2}(rF_l)= -\frac{2m}{\hbar^2}\biggl\{\! E\!+\!\frac{e^2}{r}\!-\!\frac{l(l\!+\!1)\hbar^2}{2mr^2} \!\biggr\}F_l. \end{equation} A mysterious term has been added to the potential energy. Although we got this term by some mathematical shenanigan, it has a simple physical origin. We can give you an idea about where it comes from in terms of a semi-classical argument. Then perhaps you will not find it quite so mysterious.  Think of a classical particle moving around some center of force. The total energy is conserved and is the sum of the potential and kinetic energies \begin{equation*} U = V(r) + \tfrac{1}{2}mv^2 = \text{constant}. \end{equation*} In general, $v$ can be resolved into a radial component $v_r$ and a tangential component $r\dot{\theta}$; then \begin{equation*} v^2=v_r^2+(r\dot{\theta})^2. \end{equation*} Now the angular momentum $mr^2\dot{\theta}$ is also conserved; say it is equal to $L$. We can then write \begin{equation*} mr^2\dot{\theta}=L,\quad \text{or}\quad r\dot{\theta}=\frac{L}{mr}, \end{equation*} and the energy is \begin{equation*} U = \tfrac{1}{2}mv_r^2+V(r)+\frac{L^2}{2mr^2}. \end{equation*} If there were no angular momentum we would have just the first two terms. Adding the angular momentum $L$ does to the energy just what adding a term $L^2/2mr^2$ to the potential energy would do. But this is almost exactly the extra term in (19.46). The only difference is that $l(l+1)\hbar^2$ appears for the angular momentum instead of $l^2\hbar^2$ as we might expect. But we have seen before (for example, Volume II, Section 34-7) that this is just the substitution that is usually required to make a quasi-classical argument agree with a correct quantum-mechanical calculation. We can, then, understand the new term as a “pseudo-potential” which gives the “centrifugal force” term that appears in the equations of radial motion for a rotating system. (See the discussion of “pseudo-forces” in Volume I, Section 12-5.)  We are now ready to solve Eq. (19.46) for $F_l(r)$. It is very much like Eq. (19.8), so the same technique will work again. Everything goes as before until you get to Eq. (19.19) which will have the additional term \begin{equation} \label{Eq:III:19:47} -l(l+1)\sum_{k=1}^\infty a_k\rho^{k-2}. \end{equation} This term can also be written as \begin{equation} \label{Eq:III:19:48} -l(l+1)\biggl\{\frac{a_1}{\rho}+ \sum_{k=1}^\infty a_{k+1}\rho^{k-1}\biggr\}. \end{equation} (We have taken out the first term and then shifted the running index $k$ down by $1$.) Instead of Eq. (19.20) we have  \begin{equation} \begin{aligned} \sum_{k=1}^\infty&[\{k(k+1)-l(l+1)\}a_{k+1}-2(\alpha k-1)a_k]\rho^{k-1}\\ &-\frac{l(l+1)a_1}{\rho}=0. \end{aligned} \label{Eq:III:19:49} \end{equation}   There is only one term in $\rho^{-1}$, so it must be zero. The coefficient $a_1$ must be zero (unless $l=0$ and we have our previous solution). Each of the other terms is made zero by having the square bracket come out zero for every $k$. This condition replaces Eq. (19.22) by \begin{equation} \label{Eq:III:19:50} a_{k+1}=\frac{2(\alpha k-1)}{k(k+1)-l(l+1)}\,a_k. \end{equation} This is the only significant change from the spherically symmetric case.  As before the series must terminate if we are to have solutions which can represent bound electrons. The series will end at $k=n$ if $\alpha n=1$. We get again the same condition on $\alpha$, that it must be equal to $1/n$, where $n$ is some positive integer. However, Eq. (19.50) also gives a new restriction. The index $k$ cannot be equal to $l$, the denominator becomes zero and $a_{l+1}$ is infinite. That is, since $a_1=0$, Eq. (19.50) implies that all successive $a_k$ are zero until we get to $a_{l+1}$, which can be nonzero. This means that $k$ must start at $l+1$ and end at $n$.  Our final result is that for any $l$ there are many possible solutions which we can call $F_{n,l}$ where $n\geq l+1$. Each solution has the energy \begin{equation} \label{Eq:III:19:51} E_n=-\frac{me^4}{2\hbar^2}\biggl(\frac{1}{n^2}\biggr). \end{equation} The wave function for the state of this energy with the angular quantum numbers $l$ and $m$ is \begin{equation} \label{Eq:III:19:52} \psi_{n,l,m}=a\,Y_{l,m}(\theta,\phi)F_{n,l}(\rho), \end{equation} with \begin{equation} \label{Eq:III:19:53} \rho F_{n,l}(\rho)=e^{-\alpha\rho}\sum_{k=l+1}^na_k\rho^k. \end{equation} The coefficients $a_k$ are obtained from (19.50). We have, finally, a complete description of the states of a hydrogen atom. 