If the solution of the differential equation

$${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$$ satisfies $y(0) = 0$, then the value of y(2) is _______________.

<p>$\because$ ${{dy} \over {dx}} + {e^x}({x^2} - 2)y = ({x^2} - 2x)({x^2} - 2){e^{2x}}$</p> <p>Here, $I.F. = {e^{\int {{e^x}({x^2} - 2)dx} }}$</p> <p>$= {e^{({x^2} - 2x){e^x}}}$</p> <p>$\therefore$ Solution of the differential equation is</p> <p>$$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {({x^2} - 2x)({x^2} - 2){e^{2x}}\,.\,{e^{({x^2} - 2x){e^x}}}dx} $$</p> <p>$= \int {({x^2} - 2x){e^x}\,.\,({x^2} - 2){e^x}\,.\,{e^{({x^2} - 2x){e^x}}}dx}$</p> <p>Let $({x^2} - 2x){e^x} = t$</p> <p>$\therefore$ $({x^2} - 2){e^x}dx = dt$</p> <p>$y\,.\,{e^{({x^2} - 2x){e^x}}} = \int {t\,.\,{e^t}dt}$</p> <p>$y\,.\,{e^{({x^2} - 2x){e^x}}} = ({x^2} - 2x - 1){e^{({x^2} - 2x){e^x}}} + c$</p> <p>$\therefore$ $y(0) = 0$</p> <p>$\therefore$ $c = 1$</p> <p>$\therefore$ $y = ({x^2} - 2x - 1) + {e^{(2x - {x^2}){e^x}}}$</p> <p>$\therefore$ $y(2) = - 1 + 1 = 0$</p>