If the curve, $$y = y(x)$$ represented by the solution of the differential equation $$(2xy^2 - y)dx + x \, dy = 0$$, passes through the intersection of the lines, $$2x - 3y = 1$$ and $$3x + 2y = 8$$, then $$|y(1)|$$ is equal to ______.

The differential equation is $$(2xy^2 - y)\,dx + x\,dy = 0$$. We rewrite this as $$2xy^2\,dx + (x\,dy - y\,dx) = 0$$.

Dividing through by $$x^2$$: $$\frac{2y^2}{x}\,dx + \frac{x\,dy - y\,dx}{x^2} = 0$$. Noting that $$d\!\left(\frac{y}{x}\right) = \frac{x\,dy - y\,dx}{x^2}$$, we get $$\frac{2y^2}{x}\,dx + d\!\left(\frac{y}{x}\right) = 0$$.

Let $$v = \frac{y}{x}$$, so $$y = vx$$. Then $$\frac{2v^2x^2}{x}\,dx + dv = 0$$, which gives $$2v^2 x\,dx + dv = 0$$. This is separable: $$\frac{dv}{v^2} = -2x\,dx$$.

Integrating both sides: $$-\frac{1}{v} = -x^2 + C$$, so $$\frac{1}{v} = x^2 - C$$, which means $$\frac{x}{y} = x^2 - C$$.

The curve passes through the intersection of $$2x - 3y = 1$$ and $$3x + 2y = 8$$. Solving: from the first equation $$x = \frac{1+3y}{2}$$, substituting into the second: $$\frac{3(1+3y)}{2} + 2y = 8$$, giving $$3 + 9y + 4y = 16$$, so $$13y = 13$$, hence $$y = 1$$ and $$x = 2$$.

Substituting $$(2, 1)$$: $$\frac{2}{1} = 4 - C$$, so $$C = 2$$. The solution curve is $$\frac{x}{y} = x^2 - 2$$.

At $$x = 1$$: $$\frac{1}{y} = 1 - 2 = -1$$, giving $$y = -1$$. Therefore, $$|y(1)| = 1$$.