$$\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}+x y=5 x^1 \sqrt{1+x^2} \\ & \frac{d y}{d x}+\frac{x y}{1+x^2}=\frac{5 x^2}{\sqrt{1+x^2}} \\ & \therefore \text { I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+x^2} \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & \therefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \\ & y \sqrt{1+x^2}=\frac{5 x^3}{3}+C \\ & \because y(0)=0 \Rightarrow 0=0+C \Rightarrow C=0 \\ & \therefore y=\frac{5 x^3}{3 \sqrt{1+x^2}} \\ & y(\sqrt{3})=\frac{15 \sqrt{3}}{32}=\frac{5 \sqrt{3}}{2} \end{aligned}$$