$$\because$$ $${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$$

Here, $$I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}$$

$$ = {e^{({x^2} - 2x){e^x}}}$$

$$\therefore$$ Solution of the differential equation is

$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$

$$ = \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$

Let $$({x^2} - 2x){e^x} = t$$

$$\therefore$$ $$({x^2} - 2){e^x}dx = dt$$

$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt} $$

$$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$$

$$\therefore$$ $$y(0) = 0$$

$$\therefore$$ $$c = 1$$

$$\therefore$$ $$y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$$

$$\therefore$$ $$y(2) = - 1 + 1 = 0$$