If the curve y=y(x)represented by the solution of the differential equation (2xy2−y)dx+xdy =0, passes through the intersection of the lines 2x – 3y =1 and 3x + 2y =8 then |y(1)| is equal to………..

(2xy2−y)dx+xdy =0⇒2xy2dx=  ydx−xdy⇒ 2xdx=ydx −xdyy2Integrate both sides, we getx2=xy+CThe above curve is passing through the point of intersection of two lines 2x – 3y =1 and 3x + 2y =8, it is 2,1Hence,4=2+C⇒C=2The curve is x2=xy+2 , to get the required value plug in x=1Therefore, 1=1y+21y=−1y=−1y=1