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*}
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 8 - Mathematical Modeling With Differential Equations - 8.2 Separation Of Variables - Exercises Set 8.2 - Page 575: 6
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