Let y be a solution to the given differential equation,
[tex]y' = xy[/tex]
where
[tex]\displaystyle y = \sum_{n=0}^\infty c_n x^n \\\\ y' = \sum_{n=0}^\infty nc_nx^{n-1} = \sum_{n=1}^\infty nc_nx^{n-1} = \sum_{n=0}^\infty (n+1)c_{n+1}x^n[/tex]
Substituting these series into the DE gives
[tex]\displaystyle \sum_{n=0}^\infty (n+1)c_{n+1}x^n = x\sum_{n=0}^\infty c_nx^n \\\\ \sum_{n=0}^\infty (n+1)c_{n+1}x^n = \sum_{n=0}^\infty c_nx^{n+1} \\\\ \sum_{n=0}^\infty (n+1)c_{n+1}x^n = \sum_{n=1}^\infty c_{n-1}x^n \\\\ c_1 + \sum_{n=1}^\infty (n+1)c_{n+1}x^n = \sum_{n=1}^\infty c_{n-1}x^n \\\\ c_1 + \sum_{n=1}^\infty \bigg((n+1)c_{n+1}-c_{n-1}\bigg)x^n = 0[/tex]
Then the coefficients [tex]c_n[/tex] in the series solution are governed by the recurrence,
[tex]\begin{cases}c_0 = c_0 \\ c_1 = 0 \\ (n+1)c_{n+1}-c_{n-1} = 0&\text{for }n\ge1\end{cases}[/tex]
We have
[tex](n+1)c_{n+1}-c_{n-1} = 0 \implies nc_n - c_{n-2} = 0 \implies c_n = \dfrac{c_{n-2}}n[/tex]
so it follows that [tex]c_1=c_3=c_5=\cdots = 0[/tex], while
[tex]c_0 = \dfrac{c_0}1 = \dfrac{c_0}{2^0\times0!} \\\\ c_2 = \dfrac{c_0}2 = \dfrac{c_0}{2^1\times1!}\\\\ c_4 = \dfrac{c_2}4 = \dfrac{c_0}{2\times4} = \dfrac{c_0}{2^2\times2!}\\\\ c_6 = \dfrac{c_4}6 = \dfrac{c_0}{2\times4\times6} = \dfrac{c_0}{2^3\times3!}[/tex]
and so on, with the general n-th coefficient being
[tex]c_n = \begin{cases}0&\text{if }n\text{ is odd} \\ \dfrac{c_0}{2^{n/2}\left(\frac n2\right)!} &\text{if }n\text{ is even}\end{cases}[/tex]
Then the power series solution is
[tex]\displaystyle y(x) = c_0 \sum_{n=0}^\infty \frac{x^n}{2^{n/2}\left(\frac n2\right)!} = c_0 \sum_{n=0}^\infty \frac1{\left(\frac n2\right)!} \left(\frac x{\sqrt2}\right)^n[/tex]
but this doesn't tell the whole story because it doesn't capture the odd-index-is-zero case.
More concisely: let n = 2k for integers k ≥ 0. Then
[tex]\displaystyle y(x) = c_0 \sum_{k=0}^\infty \frac{x^{2k}}{2^k k!} = c_0 \sum_{k=0}^\infty \frac1{k!} \left(\frac{x^2}2\right)^k[/tex]
and as a bonus, it's easier to get an exact solution for this DE,
[tex]y(x) = c_0e^{x^2/2}[/tex]
