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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, by hand, 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 obtain using Mathematica routine. Any help is appreciated.

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For finding asymptotic solutions you can use AsymptoticDSolveValue function.

eqn = (y^2 + 1) R''[y] + y (2 - p (b0 Sqrt[y^2 + 1])^(-p)) R'[y] - l (l + 1) R[y]
AsymptoticDSolveValue[eqn == 0, R[y], {y, \[Infinity], 1}]

Unfortunately this does not work for arbitrary p, thus you will not get a general answer. However, you can just try it for different values of p, sometimes it gives result, sometimes not, e.g.,

AsymptoticDSolveValue[eqn == 0 /. {p -> 5/2}, R[y], {y, \[Infinity], 1}]

I hope that helps.

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  • $\begingroup$ Even AsymptoticDSolveValue[eqn == 0 /. {p -> 5/2}, R[y], {y, [Infinity], 2}] does not work. $\endgroup$ – user64494 Mar 28 at 5:40
  • $\begingroup$ Yes, I know. And maybe that answers the question? I am not sure how serious is the implementation of AsymptoticDSolveValue, but if it cannot handle that, it might be an indication that it might be difficult. Similarly, if Integrate cannot produce any result it is an indication that the integral might be difficult to calculate. But again, this function has been introduced in MMA 11.3, so maybe it is not so thoroughly developed as Integrate. $\endgroup$ – MK. Mar 28 at 9:39
  • $\begingroup$ Hi @MK. Thanks for this. Even in my own routine, I cannot find a solution for non integer positive values of p. What I found is that in these cases is that if we try to expand the ODE and solve for the individual terms, there are two separate terms that solve the same coefficient. So i think some of the coefficients (e.g. $a_2$) cannot be solved uniquely. $\endgroup$ – user583893 Mar 29 at 2:04

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