Answer
$y=C_{0} e^{x^2/2}$
Work Step by Step
Here, we have $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
and $y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
Since we are given that $y' =xy$, then we have
$\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x[ \Sigma^\infty_{n=0}C_{n}x^{n}]$
This implies that $c_1+\Sigma^\infty_{n=0}[(n+2)c_{n+2} x^{n+1}-c_n]x^{n+1}=0$
After simplifications, we get $c_1=0\\c_{n+2}=\dfrac{c_n}{n+2}$
Consider odd values of $n$; then, we have
$C_{3} =0 $; when $n=3$; and $C_{5} = 0$; When $n=5$;
Consider even values of $n$; then, we have
$C_{2} = \dfrac{C_0}{2}$; $C_{4} = \dfrac{C_0}{2 \times 4}$
We follow the pattern as: $C_{2n} = (\dfrac{1}{2})^n \dfrac{c_0}{n!}$
we get , $y=\Sigma^\infty_{n=0} (\dfrac{1}{2})^n \times \dfrac{c_0x^{2n}}{n!}+\Sigma^\infty_{n=0} (0)\times x^{2n+1}$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, we get $y=C_{0} e^{x^2/2}$
