Answer
$y=C_{0} e^{x^2/2}$
Work Step by Step
The equation $y' =xy$ becomes :
$\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x[ \Sigma^\infty_{n=0}C_{n}x^{n}] \implies c_1+\Sigma^\infty_{n=2} \times nc_nx^{n-1}=\Sigma^\infty_{n=0}c_{n} \times x^{n+1}$
$c_1+\Sigma^\infty_{n=0} \times (n+2)c_{n+2} x^{n+1}=\Sigma^\infty_{n=0}c_{n} \times x^{n+1}$
and $c_1+\Sigma^\infty_{n=0}[(n+2)c_{n+2} x^{n+1}-c_n]x^{n+1}=0$
$\implies c_1=0\\c_{n+2}=\dfrac{c_n}{n+2}$
Consider the odd values of $n=3,5...$, then we get
$C_{3} =0\\C_{5} = 0$;
Consider the even values of $n=2,4...$, then we get
$C_{2} = \dfrac{C_0}{2}\\ C_{4} = \dfrac{C_0}{2 \times 4}$
Thus, we have $C_{2n} = (\dfrac{1}{2})^n \dfrac{c_0}{n!}$
Thus, we get $y=\Sigma^\infty_{n=0} (\dfrac{1}{2})^n \dfrac{c_0}{n!}x^{2n}+\Sigma^\infty_{n=0} (0) x^{2n+1}$
Use: $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, we have $y=C_{0} e^{x^2/2}$
