Answer
See below
Work Step by Step
Given: $y''+2xy'+4y=0$
Let $y=\sum a_nx^n$, obtain:
$y''+2xy'+4y$$=\sum k(k-1)(k+2)a_{k}x^{k-2}+2\sum ka_{k}x^k+4\sum a_kx^k\\
=2a_2+4a_0+\sum [(k+1)(k+2)a_{k+2}+2ka_k+4a_k]x^k$
Comparing $x^k$ on both sides:
$a_2=-2a_0\\
a_{k+2}=\frac{-6k}{(k+1)(k+2)}a_k$
It gives:
$a_2=-2a_0\\
a_4=\frac{4}{3}a_0\\
a_6=-\frac{8}{15}a_0\\
a_{2k}=(-1)^k\frac{2^k}{1.3.5...(2k-1)!}\\
=(-1)^k\frac{2^{2k}k!}{(2k)!}$
and:
$a_3=a_1\\
a_4=-\frac{1}{2}a_1\\
a_7=\frac{1}{6}a_1\\
a_{2k+1}=\frac{(-1)^{k+1}}{k!}$
Therefore: $f(x)=a_0\sum \frac{2^{2k}k!}{(2k)!}x^{2k}+a_1 \sum (-1)^k \frac{x^{2k+1}}{k!}$
