Question Number 169892 by weltr last updated on 11/May/22
$$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{xe}^{−{y}} \:,\:\:{y}\left(\mathrm{1}\right)\:=\:\mathrm{0} \\ $$
Answered by mahdipoor last updated on 11/May/22
$$\Rightarrow\frac{{dy}}{{e}^{−{y}} }={e}^{{y}} {dy}=\mathrm{2}{xdx}\Rightarrow{e}^{{y}} ={x}^{\mathrm{2}} +{c} \\ $$$$\Rightarrow{x}=\mathrm{1}\:,\:{y}=\mathrm{0}\:\Rightarrow\:{c}=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow{e}^{{y}} ={x}^{\mathrm{2}} \\ $$