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I want to solve for the asymptotic solution of the following differential equation

$$ \left(y^2+1\right) R''(y)+y\left(2-p \left(b_{0} \sqrt{y^2+1}\right)^{-p}\right) R'(y)-l (l+1) R(y)=0$$

as $y\rightarrow \infty$, where $p>0$. I did the standard way by obtaining a series solution by the Frobenius method prescription in the form

$$R(y)=\sum_{n=0}^\infty \frac{a_{n}}{y^{n+k}}$$ where $k=l+1$ is the indicial exponent. I had difficulty finding for a recurrence relation for the coefficients $a_n$ for arbitrary value of the parameter $p$. Right now, I am just doing the brute force method of solving individual $a_n$ for every value of $p$. But I am just wondering whether the recurrence relation is possible to solve. Any help is appreciated.

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Hint:

Let $r=y^2+1$ ,

Then $\dfrac{dR}{dy}=\dfrac{dR}{dr}\dfrac{dr}{dy}=2y\dfrac{dR}{dr}$

$\dfrac{d^2R}{dy^2}=\dfrac{d}{dy}\left(2y\dfrac{dR}{dr}\right)=2y\dfrac{d}{dy}\left(\dfrac{dR}{dr}\right)+2\dfrac{dR}{dr}=2x\dfrac{d}{dr}\left(\dfrac{dR}{dr}\right)\dfrac{dr}{dy}+2\dfrac{dR}{dr}=2y\dfrac{d^2R}{dr^2}2y+2\dfrac{dR}{dr}=4y^2\dfrac{d^2R}{dr^2}+2\dfrac{dR}{dr}=4(r-1)\dfrac{d^2R}{dr^2}+2\dfrac{dR}{dr}$

$\therefore4r(r-1)\dfrac{d^2R}{dr^2}+2r\dfrac{dR}{dr}+2(r-1)\left(2-b_0^{-p}pr^{-\frac{p}{2}}\right)\dfrac{dR}{dr}-l(l+1)R=0$

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  • $\begingroup$ Hi @doraemonpaul. I'll check into this. Thanks for this hint. I'll get back to you in a while $\endgroup$ – user583893 Apr 23 at 0:01
  • $\begingroup$ Hi @doraemonpaul. I still can't figure it out. Can you please complete the proof? $\endgroup$ – user583893 May 7 at 14:01
  • $\begingroup$ I am having trouble with the lower bound of the two of the sums which turns out to be a function of $p$. How do I resolve this problem so that all the terms have the same lower bound? Thank you $\endgroup$ – user583893 May 18 at 14:52

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