Question

Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1−x^2}\frac{dy}{dx}+\sqrt{1−y^2}=0,âˆ\pounds xâˆ\pounds <1.$ 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, $si{n}^{−1}x+si{n}^{−1}y=c$
so, $\frac{Ï€}{6}+\frac{Ï€}{3}=c$
Hence, $si{n}^{−1}x+si{n}^{−1}y=\frac{π}{2}$
Put $x=−\frac{1}{\sqrt{2}},si{n}^{−1}y=\frac{3π}{4}$ (Not possible)