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

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Answer: A. \frac{5 \sqrt{3}}{2} \begin{aligned} & \left(1+x^2ight) \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}} \ & 	herefore 	ext { I.F. }=e^{\int \frac{x}{1+x^2} d x}=e^{\frac{\ln \left(1+x^2ight)}{2}}=\sqrt{1+x^2} \ & 	herefore y \sqrt{1+x^2}=\int \frac{5 x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} d x \ & 	herefore 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 \ & 	herefore 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}

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

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