Let y = y(x) be the solution of the differential equation $\left(\mathrm{xy}-5{\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^2}\right)\mathrm{dx}+\left(1+{\mathrm{x}}^2\right)\mathrm{dy}=0$, y(0) = 0. Then $\mathrm{y}\left(\sqrt{3}\right)$is equal to

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