Oscillating modes with quickly exponential decay tails, solved from non-linear differential equations

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Inspired by a previous post, I foresee a more generic set of real-function PDE may be solved analytically, with the property I am going to describe.

Two coupled non-linear differential equations in a radial $r$-direction in the region $r \in [0, \infty)$:

$$-a\big(\partial_r^2+\frac{\partial_r}{r}-\frac{n^2}{r^2}+c\big) U(r)+ B(r) (\partial_r-\frac{n}{r}\big) V(r)=\frac{b^2 n}{a\; c} U(r), $$ $$ -B(r) (\partial_r+\frac{n}{r}\big) U(r) + a\big(\partial_r^2+\frac{\partial_r}{r}-\frac{n^2}{r^2}+c\big) V(r) =\frac{b^2 n}{a\; c} V(r), $$ We like to solve $U(r)$ and $V(r)$.

The B(r) is given such that $B(r)$ is a nice smooth differentiable function, with $$B(0)=0$$ $$\lim_{r \to 0} B(r)=0$$ $$\lim_{r \to \infty} B(r)=b=constant >0,$$ and $B(r) \geq 0$ is monotonically increasing along $r \in [0, \infty)$, also $$a=constant >0.$$ $$c=constant \geq 0.$$

$$n \in \mathbb{Z}^+=\text{integer natural number constant} >0.$$

Both $a$, $b$ and $c$ are finite values.

I have done some analysis myself. My expected analysis find that $U(r)$ and $V(r)$ have exponential decay tails that look like $$\exp[-\int_0^r B(r')^{\#} dr']$$ The ${\#}$ means some tentative power. And both $U(r)$ and $V(r)$ likely contain Bessel functions $J_0(r),J_1(r), ...,etc$.

What are the exact solutions of $U(r)$ and $V(r)$?

I suppose that they have localized but oscillating modes along $r$, where $U(0)$ and $V(0)$ are nearly in their maximum, with exponential decay tails $\lim_{r \to 0} U(r)=\lim_{r \to 0} V(r)=0.$

If exact analytic solutions are NOT possible, please give arguments, and please feel free to take approximations. Personally, I believe that it can be solved analytically exactly by some Bessel type functions.

It will be OK, for your solution, just for focus on the simple cases that $n=1$ or $n=2$. Even though $n \in \mathbb{Z}^+$ in general.

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The $n=c=0$ case can be solved exactly in this way posted here.

(p.s. This is not a homework problem. Just try some fun trial analysis done by myself.)

Note add: There is a $(-n^2)$ minus sign added, the previous version is not quite correct.