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(A) The given differential equation is $(1+x^2)dy = (5x^2\sqrt{1+x^2} - xy)dx$.Rearranging,we get $(1+x^2)\frac{dy}{dx} + xy = 5x^2\sqrt{1+x^2}$.Divid

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$\frac{5\sqrt{3}}{2}$

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(A) The given differential equation is $(1+x^2)dy = (5x^2\sqrt{1+x^2} - xy)dx$.Rearranging,we get $(1+x^2)\frac{dy}{dx} + xy = 5x^2\sqrt{1+x^2}$.Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{x}{1+x^2}y = \frac{5x^2}{\sqrt{1+x^2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{1+x^2}$ and $Q(x) = \frac{5x^2}{\sqrt{1+x^2}}$.Integrating factor $I.F. = e^{\int P(x)dx} = e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2}\ln(1+x^2)} = \sqrt{1+x^2}$.The solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$.$y\sqrt{1+x^2} = \int \frac{5x^2}{\sqrt{1+x^2}} \cdot \sqrt{1+x^2} dx = \int 5x^2 dx = \frac{5x^3}{3} + C$.Given $y(0)=0$,we have $0\sqrt{1+0} = \frac{5(0)^3}{3} + C$,so $C=0$.Thus,$y = \frac{5x^3}{3\sqrt{1+x^2}}$.For $x=\sqrt{3}$,$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3\sqrt{1+(\sqrt{3})^2}} = \frac{5(3\sqrt{3})}{3\sqrt{4}} = \frac{15\sqrt{3}}{3(2)} = \frac{5\sqrt{3}}{2}$.

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

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