Let y=y(x) be a solution of the differential | vedclass

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(B) Given the differential equation: $\sqrt{1-x^{2}} \frac{dy}{dx} + \sqrt{1-y^{2}} = 0$.
Separating the variables,we get: $\frac{dy}{\sqrt{1-y^{2}}}

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$\frac{1}{\sqrt{2}}$

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(B) Given the differential equation: $\sqrt{1-x^{2}} \frac{dy}{dx} + \sqrt{1-y^{2}} = 0$.
Separating the variables,we get: $\frac{dy}{\sqrt{1-y^{2}}} = -\frac{dx}{\sqrt{1-x^{2}}}$.
Integrating both sides: $\int \frac{dy}{\sqrt{1-y^{2}}} = -\int \frac{dx}{\sqrt{1-x^{2}}}$.
This gives: $\sin^{-1} y = -\sin^{-1} x + C$,or $\sin^{-1} x + \sin^{-1} y = C$.
Given $y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$,substitute $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$:
$\sin^{-1}\left(\frac{1}{2}\right) + \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = C$.
$\frac{\pi}{6} + \frac{\pi}{3} = C \implies C = \frac{\pi}{2}$.
So,the equation is $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\sin^{-1} y = \cos^{-1} x$,which implies $y = \sin(\cos^{-1} x) = \sqrt{1-x^2}$.
Now,to find $y\left(-\frac{1}{\sqrt{2}}\right)$,substitute $x = -\frac{1}{\sqrt{2}}$:
$y = \sqrt{1 - \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

Let $y=y(x)$ be a solution of the differential equation,$\sqrt{1-x^{2}} \frac{dy}{dx}+\sqrt{1-y^{2}}=0, |x| < 1$. If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$,then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to

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