A solution curve of the differential equation x2+xy+4x+2y+4dydx-y2=0,x>0, passes through the point 1,3. Then the solution curve is

Given equation is x+22+yx+2dydx-y2=0  Let x+2=X, y=YX2+XYdY−Y2dX=0dYdX=Y2X2+XYY=vX⇒dvdX+v=v2v+1             ⇒XdvdX=−vv+1          ⇒ v+1vdv=−1XdX         ⇒v+logv=−logX+logc ⇒YX+logY=logc⇒ yx+2+logy=logcit passes through 1,3⇒1+log3=logctherefore yx+2+logy=1+log3option 1: if y=x+2 then x+2x+2+logy=1+log3    logy=log3  ⇒y=3 hence 1,3 is common solutionoption3 : if y=x+22 then x+22x+2+logx+22=1+log3⇒x+2+logx+22=1+log3, no value of x satisfies.option 4: if y=x+32 then x+32x+2+logx+32=1+log3,no value of x satisfies.