Bessel ODE 2
The ordinary differential equation by Bessel is defined by:
$$
x^2\, y_\lambda'' + x\, y_\lambda' + (x^2 - \lambda^2)\, y_\lambda = 0
$$
Everything in the real numbers. With operator calculus it reads:
$$
\left[x^2\frac{d^2}{dx^2} + x \frac{d}{dx} + x^2-\lambda^2 \right]\, y_\lambda(x) = 0
\\
\left[\left(\frac{d}{dx}\right)^2 + \frac{1}{x} \frac{d}{dx}
- \left(\frac{\lambda}{x}\right)^2 \right]\, y_\lambda(x) = -1\; y_\lambda(x)
$$
Let's try to decompose the operator between $\left[\,\right]$ in factors:
$$
\left[\frac{d}{dx}+\frac{a}{x}\right]\left[\frac{d}{dx}+\frac{b}{x}\right]=
\left(\frac{d}{dx}\right)^2 + \frac{a+b}{x}\frac{d}{dx} + \frac{ab-b}{x^2}
$$
Which is equal to the operator between $\left[\,\right]$ iff:
$$
\begin{matrix} b=1-a & , & b(1-a)=\lambda^2 \end{matrix}
\quad \Longrightarrow \quad
\left\{ \begin{matrix} b = \lambda & , & a=1-\lambda \\
b = -\lambda & , & a=1+\lambda \end{matrix} \right.
$$
So there are two decompositions:
$$
\left(\frac{d}{dx}\right)^2 + \frac{1}{x} \frac{d}{dx}
- \left(\frac{\lambda}{x}\right)^2 = \left\{ \Large \begin{matrix}
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right]
\left[\frac{d}{dx}+\frac{\lambda}{x}\right] \\
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]
\left[\frac{d}{dx}-\frac{\lambda}{x}\right] \end{matrix} \right.
$$
In the first decomposition, replace $\lambda$ by $(\lambda+1)$
and write the corresponding equation:
$$
\left[\frac{d}{dx}-\frac{\lambda}{x}\right]
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]y_{\lambda+1}(x) = -1\,y_{\lambda+1}(x)
$$
Multiply the equation with $\left[d/dx+(\lambda+1)/x\right]$ from the left
at both sides of the equality sign:
$$
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]\left[\frac{d}{dx}-\frac{\lambda}{x}\right]
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]y_{\lambda+1}(x) =
-1\,\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]y_{\lambda+1}(x)
$$
And compare the outcome with the equation corresponding with the second decomposition:
$$
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]
\left[\frac{d}{dx}-\frac{\lambda}{x}\right] y_\lambda(x) = -1\, y_\lambda(x)
$$
Then we see that:
$$
\left[\frac{d}{dx}+\frac{\lambda+1}{x}\right]y_{\lambda+1}(x)=y_\lambda(x)
$$
Replace $\lambda$ by $(\lambda-1)$ herein, then we have:
$$
y_{\lambda-1}(x)=\left[\frac{d}{dx}+\frac{\lambda}{x}\right]y_{\lambda}(x)
$$
Now replace $\lambda$ by $(\lambda-1)$ in the equation corresponding
with the second decomposition:
$$
\left[\frac{d}{dx}+\frac{\lambda}{x}\right]
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right] y_{\lambda-1}(x) = -1\,y_{\lambda-1}(x)
$$
Multiply the equation with $\left[d/dx-(\lambda-1)/x\right]$ from the left
at both sides of the equality sign:
$$
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right]\left[\frac{d}{dx}+\frac{\lambda}{x}\right]
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right] y_{\lambda-1}(x) =
-1\,\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right]y_{\lambda-1}(x)
$$
And compare the outcome with the equation corresponding with the first decomposition:
$$
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right]
\left[\frac{d}{dx}+\frac{\lambda}{x}\right]y_{\lambda}(x)=-1\,y_{\lambda}(x)
$$
Then we see that:
$$
\left[\frac{d}{dx}-\frac{\lambda-1}{x}\right]y_{\lambda-1}(x)=y_\lambda(x)
$$
Replace $\lambda$ by $(\lambda+1)$ herein, then we have:
$$
y_{\lambda+1}(x)=\left[\frac{d}{dx}-\frac{\lambda}{x}\right]y_{\lambda}(x)
$$
In summary, we have found the so-called ladder relations
for solutions of the Bessel equation:
$$
y_{\lambda-1}(x)=\left[\frac{d}{dx}+\frac{\lambda}{x}\right]y_{\lambda}(x) \\
y_{\lambda+1}(x)=\left[\frac{d}{dx}-\frac{\lambda}{x}\right]y_{\lambda}(x)
$$
Suppose that we have found a solution $y_{\lambda}(x)$ for a certain value of $\lambda$.
Then the ladder relations enable us to find new solutions $y_{\lambda-1}(x)$ and
$y_{\lambda+1}(x)$ for values $\lambda\pm 1$ and from these many other solutions:
$y_{\lambda\pm 2}(x)$ , $y_{\lambda\pm 3}(x)$ and so on and so forth. Starting
with an initial value of $\lambda$ a whole range of functions is obtained in this
way, which is called the ladder of $\lambda$.