#### Answer

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#### Work Step by Step

Since $ y=25e^{-2x^2}$, then
$$ y'=-100xe^{-2x^2}.$$
Now, by substitution of $ y $ and $ y'$ into the equation, we verify that $y$ is a solution of the given differential equation.

Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

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