Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 6

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

Need to apply product rule of differentiation. This implies, $y'=e^{-x^2}+(-2xe^{-x^2})(x-2)$ As are given that $y=(x-2) e^{-x^2}$ This implies, $y'=e^{-x^2}-2xy$ Now, apply the initial conditions. $y(2)=(2-2) (e^{-2^2})=0$