If the curve , y = y(x) represented by the solution of the differential equation `(2xy^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 _______ .

Answer

To solve the given problem, we need to follow these steps:

### Step 1: Solve the Differential Equation
The differential equation given is:
\[ (2xy^2 - y)dx + xdy = 0 \]
We can rearrange this equation:
\[ 2xy^2 dx + x dy - y dx = 0 \]
This can be rewritten as:
\[ 2xy^2 dx = y dx - x dy \]
Now, we can divide both sides by \(y^2\) (assuming \(y \neq 0\)):
\[ 2x dx = \frac{y dx - x dy}{y^2} \]

### Step 2: Integrate Both Sides
Now we can integrate both sides. The left side integrates to:
\[ \int 2x \, dx = x^2 + C \]
For the right side, we need to integrate:
\[ \int \frac{y dx - x dy}{y^2} \]
This can be simplified to:
\[ \int \left( \frac{1}{y} dx - \frac{x}{y^2} dy \right) \]
The integration yields:
\[ \frac{x}{y} + C \]

### Step 3: Set Up the Equation
Combining both results, we get:
\[ x^2 = \frac{x}{y} + C \]
Rearranging gives:
\[ xy = x + Cy \]
Thus, we can express \(C\) as:
\[ xy - x = Cy \]

### Step 4: Find the Intersection of the Lines
Next, we need to find the intersection of the lines:
1. \(2x - 3y = 1\)
2. \(3x + 2y = 8\)

To find the intersection, we can solve these equations simultaneously. From the first equation, we can express \(y\):
\[ y = \frac{2x - 1}{3} \]
Substituting this into the second equation:
\[ 3x + 2\left(\frac{2x - 1}{3}\right) = 8 \]
Multiplying through by 3 to eliminate the fraction:
\[ 9x + 4x - 2 = 24 \]
Combining like terms:
\[ 13x = 26 \implies x = 2 \]
Substituting \(x = 2\) back into the first equation to find \(y\):
\[ 2(2) - 3y = 1 \implies 4 - 3y = 1 \implies 3y = 3 \implies y = 1 \]
Thus, the intersection point is \((2, 1)\).

### Step 5: Substitute the Intersection Point into the Curve Equation
Now, we substitute the point \((2, 1)\) into the equation \(xy = x + C\) to find \(C\):
\[ 2(1) = 2 + C \implies 2 = 2 + C \implies C = 0 \]
Thus, the equation of the curve simplifies to:
\[ xy = x \]

### Step 6: Find \(y(1)\)
Now we need to find \(y(1)\):
\[ 1y = 1 \implies y = 1 \]
Thus, \(|y(1)| = |1| = 1\).

### Final Answer
The final answer is:
\[ \boxed{1} \]