Answer
$y=C_{0} e^{x^2/2}$
Work Step by Step
Given: $y' =xy$
Since, $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
Here, we have
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
Then, the given equation $y' =xy$ becomes :
$\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x[ \Sigma^\infty_{n=0}C_{n}x^{n}]$
or, $c_1+\Sigma^\infty_{n=2}nc_nx^{n-1}=\Sigma^\infty_{n=0}c_{n}x^{n+1}$
or, $c_1+\Sigma^\infty_{n=0}(n+2)c_{n+2} x^{n+1}=\Sigma^\infty_{n=0}c_{n}x^{n+1}$
or, $c_1+\Sigma^\infty_{n=0}[(n+2)c_{n+2} x^{n+1}-c_n]x^{n+1}=0$
This gives: $c_1=0$ and $c_{n+2}=\dfrac{c_n}{n+2}$
For odd values of $n=3,5...$, we have
$C_{3} =0$; and $C_{5} = 0$;
For even values of $n=2,4...$, we have
$C_{2} = \dfrac{C_0}{2}$; $C_{4} = \dfrac{C_0}{2 \times 4}$
This gives the pattern $C_{2n} = (\dfrac{1}{2})^n \dfrac{c_0}{n!}$
Thus, $y=\Sigma^\infty_{n=0} (\dfrac{1}{2})^n \dfrac{c_0}{n!}x^{2n}+\Sigma^\infty_{n=0} (0) x^{2n+1}$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, $y=C_{0} e^{x^2/2}$
