## University Calculus: Early Transcendentals (3rd Edition)

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