1. Find the point through which the solution curve passes

The curve passes through the intersection of $2x-3y=1$ and $3x+2y=8.$

Solve these simultaneously.

From $2x-3y=1 \,\quad ...(1)$ $3x+2y=8 \,\quad ...(2)$

Multiply (1) by $2$: $4x-6y=2$

Multiply (2) by $3$: $9x+6y=24$

Add: $13x=26 \Rightarrow x=2$

Substitute into (2): $3(2)+2y=8 \Rightarrow 6+2y=8 \Rightarrow y=1$

So the curve passes through $(2,1).$

2. Solve the differential equation

Given: $(2xy^2-y)dx+xdy=0$

Rewrite in differential form: $x\frac{dy}{dx}=-(2xy^2-y)=y-2xy^2$

So, $\frac{dy}{dx}=\frac{y}{x}-2y^2$

This is convenient if we use the substitution $v=\frac{1}{y}.$

Then $\frac{dv}{dx}=-\frac{1}{y^2}\frac{dy}{dx}.$

Using $\frac{dy}{dx}=\frac{y}{x}-2y^2,$ we get

Since $v=\frac{1}{y}$, $\frac{dv}{dx}+\frac{1}{x}v=2.$

This is a linear differential equation.

3. Solve the linear equation

$\frac{dv}{dx}+\frac{1}{x}v=2$

Integrating factor: $\text{I.F.}=e^{\int \frac{1}{x}dx}=e^{\ln x}=x.$

Multiplying throughout by $x$: $x\frac{dv}{dx}+v=2x$

Left side is: $\frac{d}{dx}(xv)=2x$

Integrate: $xv=x^2+C$

So, $v=x+\frac{C}{x}.$

Since $v=\frac{1}{y}$, \frac{1}{y}=x+\frac{C}{x}=rac{x^2+C}{x}.

Hence, $y=\frac{x}{x^2+C}.$

4. Use the initial condition $(2,1)$

Substitute $x=2$, $y=1$: $1=\frac{2}{4+C}$

So, $4+C=2 \Rightarrow C=-2.$

Thus the particular solution is $y=\frac{x}{x^2-2}.$

5. Find $y(1)$

$y(1)=\frac{1}{1-2}=-1.$

Therefore, $|y(1)|=1.$

6. Compare with stored answer

Derived answer: $1$

Stored correct answer: $1$

They agree.