JEE Main 2025 — 24 January, Morning Shift — Question 9

$\left(1+x^{2}\right) \frac{d y}{d x}+x y=5 x^{2} \sqrt{1+x^{2}}$

$\frac{d y}{d x}+\frac{x y}{1+x^{2}}=\frac{5 x^{2}}{\sqrt{1+x^{2}}}$

$\therefore$ I.F. $=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\frac{\ln \left(1+\mathrm{x}^{2}\right)}{2}}=\sqrt{1+\mathrm{x}^{2}}$

$\therefore \mathrm{y} \sqrt{1+\mathrm{x}^{2}}=\int \frac{5 \mathrm{x}^{2}}{\sqrt{1+\mathrm{x}^{2}}} \cdot \sqrt{1+\mathrm{x}^{2}} \mathrm{dx}$

$y \sqrt{1+x^{2}}=\frac{5 x^{3}}{3}+C$

$\because \mathrm{y}(0)=0 \Rightarrow 0=0+\mathrm{C} \Rightarrow \mathrm{C}=0$

$\therefore \mathrm{y}=\frac{5 \mathrm{x}^{3}}{3 \sqrt{1+\mathrm{x}^{2}}}$

$\mathrm{y}(\sqrt{3})=\frac{15 \sqrt{3}}{3.2}=\frac{5 \sqrt{3}}{2}$