# Convert the power series solution of $(1+x^2)y''+4xy'+2y=0$ into simple closed-form expression

$(a)$Use two power series in $x$ to find the general solution of $$(1+x^2)y''+4xy'+2y=0$$

and state the set of $x$-values on which each series solution is valid.

$(b)$ Convert the power series solutions in $(a)$ into simple closed-form expressions.

$(c)$ Use $(b)$ to find the general solution of the equation above on the whole real line.

For $(a)$, I used $y(x)=\sum\limits_{n=0}^\infty a_nx^n$,and solved the recurrence relation to be $a_{2k} = (-1)^ka_0$ and $a_{2k+1} = (-1)^ka_1$, where $a_0$ and $a_1$ are arbitrary.

And for the solution I got $y(x)=a_0\sum\limits_{k=1}^\infty (-1)^kx^{2k} +a_1\sum\limits_{k=1}^\infty (-1)^kx^{2k+1}$.

How do I convert the solution into simple closed-from expressions, and how should I solve $(c)$?

Thanks for any help.

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The terms (in both) each have a ratio of $(-x^2)$ –  Aaron Jan 20 at 6:56
I got $y(x) = \dfrac{a_0}{1+x^2} +\dfrac{a_1x}{1+x^2}$, is that right? And what does it mean by "general solution on the whole real line"? –  user59036 Jan 21 at 0:27
That would be right if your sums started with $k=0$, but you've written that they start from $k=1$ (which I think might be a typo). Note that the power series only converge for $\left| x \right| < 1$ (plus whatever happens at the boundary), s0 it does not define a solution everywhere. A pirori, you only know that the function you've found satisfies the differential equation in that region. Show that it actually works everywhere, either by using a theorem or just plugging it back into the differential equation. –  Aaron Jan 21 at 0:36