Question Number 169553 by ali009 last updated on 02/May/22
$${solve}\:{the}\:{D}.{E} \\ $$$$\mathrm{2}{dx}−{e}^{{y}−{x}} {dy}=\mathrm{0} \\ $$
Answered by som(math1967) last updated on 03/May/22
$$\:\mathrm{2}{dx}={e}^{{y}−{x}} {dy} \\ $$$$\mathrm{2}{dx}=\frac{{e}^{{y}} {dy}}{{e}^{{x}} } \\ $$$$\mathrm{2}\:\int{e}^{{x}} {dx}=\int{e}^{{y}} {dy} \\ $$$$\mathrm{2}{e}^{{x}} ={e}^{{y}} +{C} \\ $$