Question

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

• A. $- \frac{1}{\sqrt{2}}$
• B. $- \frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $ \frac{\sqrt{3}}{2}$

Answer 1

Answer: (C)

$\frac{dy}{dx}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$ so, $\frac{dy}{\sqrt{1-y^{2}}}+\frac{dx}{\sqrt{1-x^{2}}}=0$
Integrating, $sin^{-1}x + sin^{-1}y = c$
so, $\frac{\pi}{6}+\frac{\pi}{3}=c$
Hence, $sin^{-1}x + sin^{-1}y=\frac{\pi}{2}$
Put $x=-\frac{1}{\sqrt{2}}, sin^{-1}y=\frac{3\pi}{4}$ (Not possible)