Consider the following Bessel-like equation: $$\frac{\partial^2u}{\partial \rho^2}+\frac{1}{\rho}\frac{\partial u}{\partial \rho}+u=\frac{A}{\rho}\delta(\rho) $$ Here, $\rho$ is non negative. This equation actually came from studying a particular problem in cylindrical coordinates, so that's the reasoning behind this requirement. What I have trouble figuring out (not just here but in general) is what the general method for obtaining the correct condition for the solutions at 0. I know one way is to write $u$ as a linear combination of Bessel functions etc and figure out the coefficients. But I know there is a different way, which has to do with integrating the equation, and is more general, since it also applies to other ODEs.
I tried to just integrate both sides and take a limit, but I don't know how I am supposed to integrate the delta function since $\rho$ has to be non negative.
Of course I am aware the delta is no function etc, but I'm not sure how to treat this specific problem distributionally either. I've seen that the solution is supposed to be something like: $$ \lim_{x\to0} \rho\frac{\partial u}{\partial\rho} = A $$ But I'm not sure how to obtain that here.
I've been stuck on similar problems before, which is why I would appreciate some general help and possibly some resource recommendations on how to better treat these problems.
Well, inasmuch as the object $\frac{\delta(\rho)}{\rho}$ is not a defined distribution, the ODE
$$\frac{d^2u}{d\rho^2}+\frac1\rho\frac{du}{d\rho}+u=\frac{A}{\rho}\delta(\rho)\tag1$$
is also not defined. However, we can interpret the ODE in $(1)$ to mean that in distribution
$$\begin{align} \rho\frac{d^2u}{d\rho^2}+\frac{du}{d\rho}+\rho u&=\frac{d}{d\rho}\left(\rho \frac{du}{d\rho}\right)+u\\\\ &=A\delta(\rho)\tag2 \end{align}$$
Next, let's move the singularity away from the origin and write
$$\frac{d}{d\rho}\left(\rho \frac{df}{d\rho}\right)+f=A\delta(\rho-\rho')\tag 3$$
Solution to $(3)$ is given by
$$f(\rho)=\begin{cases}BJ_0(\rho)&,\rho<\rho'\\\\CJ_0(\rho)+DY_0(\rho)&,\rho>\rho' \end{cases}$$
Enforcing continuity at $\rho=\rho'$ we must have
$$\begin{align} (B-C)J_0(\rho')=DY_0(\rho') \end{align}\tag4$$
And from $(3)$, we see that
$$\rho'\left((C-B)J_0'(\rho')+DY_0'(\rho')\right)=A \tag5$$
Solving $(4)$ and $(5)$ simultaneously, and using the Wronsian
$$J_0(x)Y_0'(x)-J_0'(x)Y_0(x)=\frac2{\pi x}$$
we find that
$$D=\frac{A\pi}{2}J_0(\rho')$$
Finally, letting $\rho'\to 0^+$, we find that
$$u(\rho)=BJ_0(\rho)+\frac{A\pi}{2}Y_0(\rho)$$