Question

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

Answer 1

Given,
$(2x{y}^2-y)dx+xdy=0$
$\Rightarrow \frac{dy}{dx}+2y^2-\frac{y}{x}=0$
$\Rightarrow -\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\left(\frac{1}{x}\right)=2$
Let, $\frac{1}{y}=z$
$-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$
$\Rightarrow \frac{dz}{dx}+z\left(\frac{1}{x}\right)=2$
I.F. $=e^{\int \frac{1}{x}dx}=x$
$\therefore z\left(x\right)=\int 2\left(x\right)dx=x^2+c$
$\Rightarrow \frac{x}{y}=x^2+c$
As it passes through P(2, 1)
[Point of intersection of $2x-3y=1$ and $3x+2y=8$]
$\therefore \frac{2}{1}=4+c$
$\Rightarrow c=-2$
$\Rightarrow \frac{x}{y}=x^2-2$
Put $x=1$
$\frac{1}{y}=1-2=-1$
$\Rightarrow y\left(1\right)=-1$
$\Rightarrow |y\left(1\right)|=1$