JEE Advanced 2025 — Paper 2 — Question 25

Put $\mathrm{y}=\mathrm{vx} \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

D.E.$\mathrm{x}^{2}\left(v+\mathrm{x} \frac{\mathrm{d} v}{\mathrm{dx}}\right)+\mathrm{x}^{2} v=\mathrm{x}^{2}\left(1+v^{2}\right)$

$\Rightarrow v+x \frac{d v}{d x}+v=1+v^{2}$

$\Rightarrow x \frac{d v}{d x}=1+v^{2}-2 v$

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

$\Rightarrow-\frac{1}{v-1}=\ln |x|+\mathrm{C}$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{x}-\mathrm{y}}=\ln |x|+\mathrm{C}=\ln \mathrm{x}+\mathrm{C} \quad\left(\right.$

Since $\left.\mathrm{x}>\frac{1}{\mathrm{e}}\right)$

Given $\mathrm{y}(1)=0$

$\Rightarrow \mathrm{C}=1$

So $\frac{x}{x-y}=\ln (e x)$

Now $y=(e)=\frac{e}{2}$ and $y\left(e^{2}\right)=\frac{2 e^{2}}{3}$

$\therefore \frac{2(y(e))^{2}}{y\left(e^{2}\right)}=\frac{2 \cdot \frac{e^{2}}{4}}{\frac{2 e^{2}}{3}}=\frac{3}{4}=0.75$