# Recurrence relation for the asymptotic expansion of an ODE

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.

Hint:

Let $$r=y^2+1$$ ,

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

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

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

• Hi @doraemonpaul. I'll check into this. Thanks for this hint. I'll get back to you in a while – user583893 Apr 23 at 0:01
• Hi @doraemonpaul. I still can't figure it out. Can you please complete the proof? – user583893 May 7 at 14:01
• 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 – user583893 May 18 at 14:52