If the curve y=y(x) represented by the solution of


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



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

$\Rightarrow \frac{d y}{d x}+2 y^{2}-\frac{y}{x}=0$

$\Rightarrow-\frac{1}{y^{2}} \frac{d y}{d x}+\frac{1}{y}\left(\frac{1}{x}\right)=2$


$-\frac{1}{y^{2}} \frac{d y}{d x}=\frac{d z}{d x}$

$\Rightarrow \quad \frac{d z}{d x}+z\left(\frac{1}{x}\right)=2$

$\Rightarrow \quad F_{x}=e^{\int \frac{1}{x} d x}=x$

$\Rightarrow z(x)=\int 2(x) d x=x^{2}+c$

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

As it passes through $\mathrm{P}(2,1)$

[Point of intersection of $2 x-3 y=1$ and $3 x+2 y=8$ ]

$\therefore \frac{2}{1}=4+c$

$\Rightarrow c=-2$

$\Rightarrow \frac{x}{y}=x^{2}-2$

Put $x=1$


$\Rightarrow y(1)=-1$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now