Q42 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics
Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to
- A$\sqrt{\frac{15}{2}}$
- Correct answerB$\frac{5 \sqrt{3}}{2}$
- C$2 \sqrt{2}$
- D$\sqrt{\frac{14}{3}}$
Solution
$\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^2 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text { I.F. }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+x^2} \mathrm{dx}}=\mathrm{e}^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+\mathrm{x}^2} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{y} \sqrt{1+\mathrm{x}^2}=\frac{5 \mathrm{x}^3}{3}+\mathrm{C} \\ & \because \mathrm{y}(0)=0 \Rightarrow 0=0+\mathrm{C} \Rightarrow \mathrm{C}=0\end{aligned}$
$\begin{aligned} & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2}\end{aligned}$
Hence, the correct answer is option: 'B'.