<s>[INST] Solving this ODE near x=0 I could not get the correct solution on this ODE:
$$(1+x^2)y''-xy'-8y=0$$ around $x=0$.
I tried to solve this using MacLaurin expansion on $y$ and forming recurrence relations between the coefficients.
Can anyone teach me how to solve this by steps? Thank you.
 [/INST] Let $y(x) = \sum^\infty_{n=0} a_n x^n$. We see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} + \sum^\infty_{n=2} n(n-1)a_nx^n - \sum^\infty_{n=1} n a_nx^n - \sum^\infty_{n=0} 8a_n x^n = 0.$$ By re-indexing and running off a few terms, we see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} = \sum^\infty_{n=0} (n+2)(n+1)a_{n+2}x^n = 2 a_2 + 6a_3x + \sum^\infty_{n=2} (n+2)(n+1)a_{n+2}x^n.$$ We also need to run terms off a couple of the other sums so that all of them are indexed starting at $n = 2$: $$\sum^\infty_{n=0} 8a_n x^n = 8a_0 + 8a_1x + \sum^\infty_{n=2} 8a_n x^n$$ and $$\sum^\infty_{n=1} n a_nx^n = a_1x + \sum^\infty_{n=2} na_nx^n.$$ Putting all this together, we see $$(2a_2 - 8a_0) + (6a_3 - 9a_1)x + \sum^\infty_{n=2} [(n+2)(n+1)a_{n+2} + (n(n-1) - n - 8)a_n ]x^n = 0$$ or $$(2a_2 - 8a_0) + (6a_3 - 9a_1)x + \sum^\infty_{n=2} [(n+2)(n+1)a_{n+2} + (n+2)(n-4)a_n ]x^n = 0.$$ Thus the coefficients satisfy $$a_2 = 4a_0, \,\,\,\,\, a_3 = \tfrac 3 2 a_1, \,\,\,\,\, a_{n+2} = -\left(\tfrac{n-4}{n+1}\right)a_n, \,\,\,\, n \ge 2.$$
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