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: $y''+xy'+y=0$
Since, $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
Here, we have
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$ and $y''=\Sigma^\infty_{n=0}(n+2) (n+1) C_{n+2} x^{n}$
Then, the given equation $y''+xy'+y=0$ becomes:
$\Sigma^\infty_{n=0}(n+2) (n+1) C_{n+2} x^{n}+x \Sigma^\infty_{n=1}nC_{n} x^{n-1}+ \Sigma^\infty_{n=0}C_{n}x^{n}=0$
or, $\Sigma^\infty_{n=1}(n+2)(n+1)c_{n+2}x^{n}+\Sigma^\infty_{n=1}n c_{n}x^{n}+C_0+\Sigma^\infty_{n=1}C_{n}x^{n}+2C_2=0$
or, $C_{n+2}(n+2)(n+1)+nC_n+C_n=0$
For n=0 , we have $C_{2} =\dfrac{-C_{0}}{2}$;
For n=1 , we have $C_{3} =\dfrac{-C_{1}}{3}$;
Thus, we get $C_{2n} = \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)}C_0$ and $C_{2n+1} = \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)}C_1$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, $y= \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)}C_0+ \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)}C_1$
or, $y=C_0 \Sigma^\infty_{n=0} \dfrac{(-1)^n}{2^n n!}x^n+C_1 \Sigma^\infty_{n=0} \dfrac{(-2)^nn!}{(2n+1)!}x^{2n+1}$
