Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1-x^{2}} \frac{d y}{d x}+\sqrt{1-y^{2}}=0,|x|<1 .$
If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$, then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to:
Correct Option: , 3
The given differential eqn. is
$\frac{d y}{\sqrt{1-y^{2}}}+\frac{d x}{\sqrt{1-x^{2}}}=0 \Rightarrow \sin ^{-1} y+\sin ^{-1} x=c$
At $x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \quad \Rightarrow \quad c=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} y=\cos ^{-1} x$
Hence, $y\left(-\frac{1}{\sqrt{2}}\right)=\sin \left(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)$
$=\sin \left(\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)=\frac{1}{\sqrt{2}}$
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