JEE Main 2024 — 27 January, Shift 2 — Question 17

$\left(x^{2}-4\right) d y-\left(y^{2}-3 y\right) d x=0$

$\Rightarrow \int \frac{\mathrm{dy}}{\mathrm{y}^{2}-3 y}=\int \frac{\mathrm{dx}}{\mathrm{x}^{2}-4}$

$\Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^{2}-4}$

$\Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C$

$\Rightarrow \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\mathrm{C}$

At $x=4, y=\frac{3}{2}$

$\therefore \mathrm{C}=\frac{1}{4} \ln 3$

$\therefore \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+\frac{1}{4} \ln (3)$

At $x=10$ $\frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3)$

$\ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0$

as $y(4)=\frac{3}{2} \Rightarrow y \in(0,3)$

$-y+3=8^{1 / 4} \cdot \mathrm{y}$

$y=\frac{3}{1+8^{1 / 4}}$