Solution of the differential equation y−xdydx=ay2+dydx

We have y−xdydx=ay2+dydx⇒ydx−xdy=ay2dx+ady⇒y(1−ay)dx=(x+a)dy⇒dxx+a−dyy(1−ay)=0 since 1y1-ay=1y+a1-ay Integrating , we getlog⁡(x+a)-log⁡y-alog⁡(1−ay)-a=log⁡clog (x+a)-log y+log(1-ay)=logc⇒log⁡(a+x)(1−ay)y=log⁡c⇒(x+a)(1−ay)=cy