If the curve $y=y\left(x\right)$represented by the solution of the differential equation $(2x{y}^2-y)dx+xdy=0$, passes through the intersection of the lines 2x – 3y =1 and 3x + 2y =8 then |y(1)| is equal to………..

$(2x{y}^2-y)dx+xdy=0\Rightarrow 2x{y}^{2}dx=ydx-xdy\Rightarrow 2xdx=\frac{ydx-xdy}{y^2}$

Integrate both sides, we get$x^2=\frac{x}{y}+C$

The above curve is passing through the point of intersection of two lines 2x – 3y =1 and 3x + 2y =8, it is $\left(2,1\right)$

Hence,$4=2+C\Rightarrow C=2$

The curve is $x^2=\frac{x}{y}+2$ , to get the required value plug in $x=1$

Therefore, 

$\begin{array}{c}1=\frac{1}{y}+2\\ \frac{1}{y}=-1\\ y=-1\\ \left|y\right|=\boxed{1}\end{array}$