*Seek a power series solution about $\displaystyle x=0$ of the equation* *$\displaystyle (1+x^2)y''+2xy'-2y=0$* *as far as the term in $\displaystyle x^6$. Then find a closed-form of the solution (i.e. write the solution using simple functions, in this case arctan, rather than infinite series) by using* *$\displaystyle \arctan{x}=x-\frac{x^3}{3}+\frac{x^5}{5}+...+(-1)^n\frac{x^{2n+1}}{2n+1}+...$.*
It's the closed-form bit that's confusing me, 'cause my answer looks nothing like the arctan thing. The answer I got was

$\displaystyle y=A(1+x^2-\frac{1}{3}x^4+\frac{2}{5}x^6+...)$ where A is just a constant.

Can anyone help?