QJEEM22Ma379b: JEE Main 2022 Mathematics - Class 12 JEEM22Ma379b Solution: If the solution of the differential equation satisfies , then…

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

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

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

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

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

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

$\therefore$ $c = 1$

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

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