If y=y(x) is a particular solution of sqrt{1- | vedclass

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(A) Given the linear differential equation: $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$. Dividing by $\sqrt{1-x^2}$,we get: $\frac{dy

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

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(A) Given the linear differential equation: $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$. Dividing by $\sqrt{1-x^2}$,we get: $\frac{dy}{dx} + \frac{2x}{1-x^2} y = \frac{x}{\sqrt{1-x^2}}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1-x^2}$ and $Q(x) = \frac{x}{\sqrt{1-x^2}}$. The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{2x}{1-x^2} dx} = e^{-\ln(1-x^2)} = \frac{1}{1-x^2}$. The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. $y \cdot \frac{1}{1-x^2} = \int \frac{x}{\sqrt{1-x^2}} \cdot \frac{1}{1-x^2} dx + C = \int x(1-x^2)^{-3/2} dx + C$. Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$. $y \cdot \frac{1}{1-x^2} = -\frac{1}{2} \int u^{-3/2} du + C = -\frac{1}{2} \frac{u^{-1/2}}{-1/2} + C = u^{-1/2} + C = (1-x^2)^{-1/2} + C$. Thus,$y = (1-x^2)^{1/2} + C(1-x^2)$. Given $y(0) = 1$,we have $1 = (1-0)^{1/2} + C(1-0) \Rightarrow 1 = 1 + C \Rightarrow C = 0$. So,$y = \sqrt{1-x^2}$. At $x = \frac{1}{2}$,$y\left(\frac{1}{2}\right) = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If $y=y(x)$ is a particular solution of $\sqrt{1-x^2} \frac{dy}{dx} + \frac{2x}{\sqrt{1-x^2}} y = x$ with $y(0)=1$,then $y\left(\frac{1}{2}\right) = $

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