# 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 ______. Option: 1 1 Option: 2 2 Option: 3 3 Option: 4 4

$\\\left(2 x y^{2}-y\right) d x+x d y=0 \\ 2 x y^{2} d x-y d x+x d y=0 \\ 2 x d x=\frac{y d x-x d y}{y^{2}}=d\left(\frac{x}{y}\right)$

now integrate

$x^2=\frac{x}{y}+c$

Now the point of intersection of lines are (2, 1)

$\\4=\frac{2}{1}+\mathrm{c} \quad \Rightarrow \mathrm{c}=2 \\ \mathrm{x}^{2}=\frac{\mathrm{x}}{\mathrm{y}}+2 \\ \text { Now } \mathrm{y}(1)=-1 \\ \Rightarrow|\mathrm{y}(1)|=1$

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