Let y= y( x) be the solution of the differential equation (x y-5 x^2 √1+x^2 ) d x+ (1+x^2 ) d y=0, y(0)=0. Then y(√3) is equal to

$\displaystyle \begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^2 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text { I.F. }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+x^2} \mathrm{dx}}=\mathrm{e}^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+\mathrm{x}^2} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \therefore \mathrm{y} \sqrt{1+\mathrm{x}^2}=\int \frac{5 \mathrm{x}^2}{\sqrt{1+\mathrm{x}^2}} \cdot \sqrt{1+\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{y} \sqrt{1+\mathrm{x}^2}=\frac{5 \mathrm{x}^3}{3}+\mathrm{C} \\ & \because \mathrm{y}(0)=0 \Rightarrow 0=0+\mathrm{C} \Rightarrow \mathrm{C}=0\end{aligned}$​(1+x2)dxdy​+xy=5x21+x2​dxdy​+1+x2xy​=1+x2​5x2​∴ I.F. =e∫1+x2x​dx=e2ln(1+x2)​=1+x2​∴y1+x2​=∫1+x2​5x2​⋅1+x2​dx∴y1+x2​=∫1+x2​5x2​⋅1+x2​dxy1+x2​=35x3​+C∵y(0)=0⇒0=0+C⇒C=0​ $\begin{aligned} & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2}\end{aligned}$​∴y=31+x2​5x3​y(3​)=32153​​=253​​​