0

Solving a differential equation with the method of series as described in this site, I arrived to this point:

$$\sum_nr^n\left[a_{n-1}(2n+n^2-3+4\rho+4n\rho+4\rho^2)+a_n(l(l+1)-2n+3-4\rho)+a_{n+1}(-l(l+1)-2n\rho-4\rho^2)\right]=0$$

how do I find the general term of the series?

If there where only two terms, I would be able to solve it as described in the above site, but there are three terms and I do not know how to proceed. For example if the series was of the type:

$$\sum_n r^n(A a_n+ B a_{n+2})$$ I would divide by A and write a table with two columns for $n$ and for $n+2$, so I would find the recursion relation, but here there is an additional term and I do not know how to consider it to solve the recurrence relation.

The differential equation is:

$$(r-1)rf^{\prime\prime}(r)+f^\prime(r)-\left[ \dfrac{\rho^2 r^3}{r-1}+l(l+1)-\dfrac{3}{r} \right]f(r)=0$$

mattiav27
  • 413
  • what is the differential equation – Mohammed M. Zerrak Feb 29 '20 at 09:36
  • since you have no $f(r)$ term in your DE, then consider $f'(r)=u$ then $f''(r)=u'$ and you'll have a first order linear DE. – Qurultay Feb 29 '20 at 10:07
  • @Qurultay sorry that was a typo the last term actually depends on $f(r)$ – mattiav27 Feb 29 '20 at 10:11
  • Could you please point out where exactly in the very long page you linked to the approach you're using is described? It seems somewhat unorthodox; I would have expected $f(r)$ to be given by a simple power series (e.g. $f(r)=\sum_n a_nr^n$) and the recurrence you get to be a sum of terms equal to $0$, not to $f(r)$. – joriki Feb 29 '20 at 11:01
  • @joriki for instance in the section with title "Technique for finding solutions near an ordinary point by the Method of Undetermined Coefficients." – mattiav27 Feb 29 '20 at 11:03
  • No, that's the standard method where you get an expression equal to $0$. Where is the approach described that resulted in your equation where $f(r)$, not $0$, is equated to a complicated sum? – joriki Feb 29 '20 at 11:06
  • @joriki I used that method descibed there, but probably I did not unsderstand it properly that's why there is the $f(r)$, I am sorry I will edit my post to eliminate that error – mattiav27 Feb 29 '20 at 11:11
  • If that series is zero, each term is zero. That gives you a recurrence. But I fear that using e.g. generating functions to solve it will lead straight back to your original differential equation. – vonbrand Feb 29 '20 at 17:47

0 Answers0