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