The correct option is A c eex22=sin(y−x)
Put y-x = z. Then dydx−1=dzdx⇒dydx=1+dzdx
Given dydx−x tan(y−x)=1⇒1+dzdx−x tan z=1⇒dzdx=x tan z⇒1tanzdz=x dx⇒∫cotz dz=∫x dx
⇒log|sin z|−logc=x22⇒log|sin zc|=x22=sinzc=ex22⇒sin(y−x)=c ex22
∴ The solution is sin(y−x)=cex22, where c is arbitrary constant.