Question

If the curve, $y=y(x)$ represented by the solution of the differential equation $\left(2x{y}^2-y\right)dx+xdy=0,$ passes through the intersection of the lines, $2x-3y=1$ and $3x+2y=8$, then $|y(1)|$ is equal to ______.

Answer 1

Answer: 1

$(2x{y}^2-y)dx+xdy=0$
$2x{y}^{2}dx-ydx+xdy=0$
$2xdx=\frac{ydx-xdy}{y^2}=d\left(\frac{x}{y}\right)$
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 $y(1)=-1$
$\Rightarrow |y(1)|=1$