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 $|y(1) |$ is equal to ______.

Answer 1

$(2xy^2 - y)dx + xdy = 0$
$2 x y^{2} d x-y d x+x d y=0$
$2 x dx =\frac{ y d x - x dy }{ y ^{2}}= d \left(\frac{ x }{ y }\right)$
Now integrate
$x ^{2}=\frac{ x }{ y }+ c$
Now point of intersection of lines are (2,1)
$4=\frac{2}{1}+ c \Rightarrow c =2$
$x^{2}=\frac{x}{y}+2$
Now $y(1)=-1$
$\Rightarrow |y(1)|=1$