Solution of the differential equation y−xdydx=ay2+dydx
x+a1−ay=cy
x−a1−ay=cy
x−a1−ay=cx
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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)-logy-alog(1−ay)-a=logclog (x+a)-log y+log(1-ay)=logc⇒log(a+x)(1−ay)y=logc⇒(x+a)(1−ay)=cy