Answer
$$y=c_1e^{\frac{-1}{2}x^2}$$
Work Step by Step
Given $$ y'=-xy $$
Separate variables
\begin{align*}
\int \frac{dy}{y} &= \int-x dx \\
\ln y&=\frac{-1}{2}x^2+c\\
y&= e^{\frac{-1}{2}x^2+c}\\
&=c_1e^{\frac{-1}{2}x^2}
\end{align*}
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