Question:
If the curve, $y=y(x)$ represented by the solution of the differential equation $\left(2 x y^{2}-y\right) d x+x d y=0$, passes through the intersection of the lines, $2 x-3 y=1$ and $3 x+2 y=8$, then $l y(1) \mid$ is equal to
Solution:
$\left(2 x y^{2}-y\right) d x+x d y=0$
$2 x y^{2} d x-y d x+x d y=0$
$2 x d x=\frac{y d x-x d y}{y^{2}}=d\left(\frac{x}{y}\right)$
Now integrate
$x^{2}=\frac{x}{y}+c$
Now point of intersection of lines are $(2,1)$
$4=\frac{2}{1}+c \quad \Rightarrow c=2$
Now $y(1)=-1$
$\Rightarrow|y(1)|=1$
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