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.
On
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 $$a_2 = 4a_0, \,\,\,\,\, a_3 = \tfrac 3 2 a_1, \,\,\,\,\, a_{n+2} = -\left(\tfrac{n-4}{n+1}\right)a_n, \,\,\,\, n \ge 2.$$
Using the ansatz \begin{align} y = \sum^\infty_{k=0} a_k x^k \end{align} then it follows \begin{align} y' = \sum^\infty_{k=1} k a_k x^{k-1} \ \ \text{ and } \ \ \ y''= \sum^\infty_{k=2} k(k-1) a_k x^{k-2}. \end{align} Plugging the power series back into the differential equation yields \begin{align} &(1+x^2)\sum^\infty_{k=2}k(k-1)a_k x^{k-2}-x\sum^\infty_{k=1}k a_k x^{k-1} -8 \sum^\infty_{k=0}a_k x^k \\ &= \sum^\infty_{k=0}(k+2)(k+1) a_{k+2}x^k +\sum^\infty_{k=2}k(k-1)a_k x^k-\sum^\infty_{k=1}ka_k x^k-8\sum^\infty_{k=0}a_k x^k\\ &= 2a_2+6a_3x+\sum^\infty_{k=2}\{(k+2)(k+1)a_{k+2}+k(k-1)a_k\}x^k -a_1 x -\sum^\infty_{k=2}ka_k x^k -8a_0-8a_1x\\ &-8\sum^\infty_{k=2}a_k x^k\\ &= (2a_2-8a_0)+(6a_3-9a_1)x+\sum^\infty_{k=2}\{(k+2)(k+1)a_{k+2}+(k^2-2k-8)a_k\}x^k=0. \end{align} So, it follows \begin{align} a_2 = 4a_0, \ a_3 = \frac{3}{2}a_1, \ a_{k+2} = -\frac{k-4}{k+1}a_k \ \text{ for } k\geq 2. \end{align} If $k$ is even then it follows \begin{align} a_{2(\ell + 1)} = -\frac{2\ell -4}{2\ell+1}a_{2\ell}= \frac{2\ell-4}{2\ell+1}\frac{2(\ell-1)-4}{2(\ell-1)+1}a_{2(\ell-1)}= (-1)^\ell\prod^{\ell}_{j=1}\frac{2j-4}{2j+1}a_2 = 4\left(\prod^{\ell}_{j=1}\frac{-2j+4}{2j+1}\right)a_0 \end{align} for $\ell \geq 1$ and likewise when $k$ is odd we have \begin{align} a_{(2\ell+1)+2} = \frac{2\ell+1 -4}{2\ell+1+1}a_{2\ell+1} = \frac{3}{2}\left(\prod^{\ell}_{j=1}\frac{-2j+3}{2j+2}\right)a_1. \end{align} Finally, we see that \begin{align} y =&\ a_0+a_1x+a_2 x^2+a_3x^3 +\sum^\infty_{k=1} a_{2k+2}x^{2k+2}+\sum^\infty_{k=1} a_{2k+3}x^{2k+3}\\ =&\ a_0\left(1+4x^2+4\sum^\infty_{k=1}\left(\prod^{k}_{j=1}\frac{-2j+4}{2j+1}\right)x^{2k+2} \right)+a_1\left(x+\frac{3}{2}x^3+\frac{3}{2}\sum^\infty_{k=1}\left(\prod^{k}_{j=1}\frac{-2j+3}{2j+2}\right)x^{2k+3} \right) \end{align}