Given,
(2xy2−y)dx+xdy=0
⇒dydx+2y2−yx=0
⇒−1y2dydx+1y(1x)=2
Let, 1y=z
−1y2dydx=dzdx
⇒dzdx+z(1x)=2
I.F. =e∫1xdx=x
∴z(x)=∫2(x)dx=x2+c
⇒xy=x2+c
As it passes through P(2, 1)
[Point of intersection of 2x−3y=1 and 3x+2y=8]
∴21=4+c
⇒c=−2
⇒xy=x2−2
Put x=1
1y=1−2=−1
⇒y(1)=−1
⇒|y(1)|=1