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**Kyeong** A. Find two linearly independent power series solutions to the homogeneous equation

(x2 + 1)y'' + 3xy' - 4y = 0

Let us start with this. We know there exists a solution to this equation $\displaystyle \Sigma_{n=0}^{\infty} a_nx^n$ (with infinite radius of convergence).

Substituting we get,

$\displaystyle (x^2+1)\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + 3x\sum_{n=1}^{\infty} na_nx^{n-1}-4\sum_{n=0}^{\infty}a_nx^n=0$.

Multiply out,

$\displaystyle \sum_{n=2}^{\infty}n(n-1)a_nx^n + \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} + 3\sum_{n=1}^{\infty} na_nx^n -4\sum_{n=0}^{\infty}a_nx^n=0$.

Change the index on the 2nd $\displaystyle \Sigma$,

$\displaystyle \sum_{n=2}^{\infty}n(n-1)a_nx^n + \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n + 3\sum_{n=1}^{\infty} na_nx^n -4\sum_{n=0}^{\infty}a_nx^n=0$.

Evaluting the 2nd sum at $\displaystyle n=0,1$ the 3rd at $\displaystyle n=1$ the 4th at $\displaystyle n=0,1$ and combining,

$\displaystyle (2a_2-4a_0)+(6a_1+3a_1-4a_1)x+\sum_{n=2}^{\infty} [n(n-1)a_n+(n+2)(n+1)a_{n+2} + 3na_n - 4a_n]x^n = 0$.

But this can be written as,

$\displaystyle \sum_{n=0}^{\infty} [n(n-1)a_n+(n+2)(n+1)a_{n+2} + 3na_n - 4a_n]x^n = 0$.

This simplifies to,

$\displaystyle \sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n^2+2n-4)a_n]x^n = 0$.

Therefore the recurrence relation is,

$\displaystyle \boxed{a_{n+2} = - \tfrac{n^2+2n-4}{n^2+3n+2}a_n} ~ \ ~ n\geq 0$.