Answer
$c_0-c_1(1-x)$
Work Step by Step
We are given that $y''+xy'+y=0$
Since, we know that $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}\\y''=\Sigma^\infty_{n=2}n(n-1) C_{n} x^{n-2}$
Now, $\Sigma^\infty_{n=2}n(n-1) C_{n} x^{n-2}+\Sigma^\infty_{n=1}n c_n x^{n-1}=0$
and $\Sigma^\infty_{n=1}n(n-1)c_{n}x^{n-1}-\Sigma^\infty_{n=1}(n+1)n c_{n+1}x^{n-1}+\Sigma^\infty_{n=1}n c_{n}x^{n-1}=0\\c_{n+1}=\dfrac{n}{n+1}c_n$
Re-arrange as: $y= \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)}c_0+ \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)}c_1$
After solving, we get $y= c_0+c_1 \Sigma^\infty_{n=1} \dfrac{x^n}{n!}=c_0-c_1(1-x)$
