University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 6


$y'=e^{-x^2}-2xy$ and $y(2)=0$

Work Step by Step

Apply the product rule of differentiation: Thus, $y'=e^{-x^2}+(-2xe^{-x^2})(x-2)$ Since we have: $y=(x-2) e^{-x^2}$ Therefore, $y'=e^{-x^2}-2xy$ Apply the initial conditions: $y(2)=(2-2) e^{-2^2}=0$
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