Answer
$C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}x^n+C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}$
Work Step by Step
Given: $y''+xy'+y=0$
$\Sigma^\infty_{n=0}(n+2) (n+1) C_{n+2} x^{n}+x \Sigma^\infty_{n=1}nC_{n} x^{n-1}+ \Sigma^\infty_{n=0}C_{n}x^{n}=0$
Re-write as: $\Sigma^\infty_{n=1}(n+2)(n+1)c_{n+2}x^{n}+\Sigma^\infty_{n=1}n c_{n}x^{n}+C_0+\Sigma^\infty_{n=1}C_{n}x^{n}+2C_2=0$
Also, we get $C_{n+2}(n+2)(n+1)+nC_n+C_n=0$
n=0 : $C_{2} =\dfrac{-C_{0}}{2}$
n=1 : $C_{3} =\dfrac{-C_{1}}{3}$
Thus, we get $C_{2n} = \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)}C_0=C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}$ and $C_{2n+1} = \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)}C_1=C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}$
Thus, we have
$y=C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}x^n+C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}$
