Question Number 37363 by math khazana by abdo last updated on 12/Jun/18
$${solve}\:{y}^{'} \:\:+{xe}^{−{x}^{\mathrm{2}} } {y}\:\:={e}^{−{x}} \:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jun/18
$$\frac{{dy}}{{dx}}+{xe}^{−{x}^{\mathrm{2}} } {y}={e}^{−{x}} \\ $$$${intregating}\:{factor} \\ $$$${e}^{\int{xe}^{−{x}^{\mathrm{2}} } {dx}} \\ $$$${t}={x}^{\mathrm{2}} \:\:\:{dt}=\mathrm{2}{xdx} \\ $$$${e}^{\int{e}^{−{t}} ×\frac{{dt}}{\mathrm{2}}\:\:} \\ $$$${e}^{\frac{{e}^{−{t}} }{−\mathrm{2}}\:\:} ={e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \\ $$$$\:{e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \frac{{dy}}{{dx}}+{xe}^{−{x}^{\mathrm{2}} } ×{e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} ×{y}={e}^{−{x}} ×{e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \\ $$$$\frac{{d}}{{dx}}\left({ye}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \right)={e}^{−{x}} ×{e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \\ $$$$\:\int{d}\left({ye}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} \right)=\int{e}^{−{x}} ×{e}^{\frac{{e}^{−{x}^{\mathrm{2}} } }{−\mathrm{2}}} {dx} \\ $$$${contd} \\ $$$$\:\:{k}={x}+\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}}\:\:\:{dk}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×{e}^{−{x}^{\mathrm{2}} } ×−\mathrm{2}{x} \\ $$$$ \\ $$