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

Solve $$(xy - 5x^2\sqrt{1+x^2})dx + (1+x^2)dy = 0$$ with y(0) = 0.

We start by rearranging the equation into a standard first‐order form. Writing $$(1+x^2)\frac{dy}{dx} + xy = 5x^2\sqrt{1+x^2}$$ and then dividing through by $$1+x^2$$ gives $$\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}\,. $$

Next, we determine the integrating factor: $$IF = e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2}\ln(1+x^2)} = \sqrt{1+x^2}\,. $$

Multiplying both sides of the differential equation by this factor yields $$\frac{d}{dx}\bigl(y\sqrt{1+x^2}\bigr) = \frac{5x^2}{\sqrt{1+x^2}}\times\sqrt{1+x^2} = 5x^2\,. $$ Integrating with respect to $$x$$ gives $$y\sqrt{1+x^2} = \int 5x^2\,dx = \frac{5x^3}{3} + C\,. $$

Applying the initial condition $$y(0)=0$$ leads to $$0\cdot1 = 0 + C\quad\Rightarrow\quad C=0\,, $$ so the solution simplifies to $$y = \frac{5x^3}{3\sqrt{1+x^2}}\,. $$

Finally, evaluating at $$x=\sqrt{3}$$ yields $$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+3}} = \frac{5\times3\sqrt{3}}{3\times2} = \frac{5\sqrt{3}}{2}\,. $$ The correct answer is Option 2: $$\frac{5\sqrt{3}}{2}\,. $$