Question

Let $y=y\left(x\right)$ 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:

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

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

$\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0$
$\Rightarrow \frac{dy}{\sqrt{1-y^2}}+\frac{dx}{\sqrt{1-x^2}}=0$
$\Rightarrow {\sin }^{-1}y+{\sin }^{-1}x=c$

If $x=\frac{1}{2},y=\frac{\sqrt{3}}{2}$
then ${\sin }^{-1}\frac{\sqrt{3}}{2}+{\sin }^{-1}\frac{1}{2}=c$
$\therefore \frac{\pi }{3}+\frac{\pi }{6}=c\Rightarrow c=\frac{\pi }{2}$

Therefore, the solution is
${\sin }^{-1}y=\frac{\pi }{2}-{\sin }^{-1}x={\cos }^{-1}x$
$\Rightarrow {\sin }^{-1}y={\cos }^{-1}x$

Now, put $x=\frac{-1}{\sqrt{2}}$
${\sin }^{-1}y={\cos }^{-1}\frac{-1}{\sqrt{2}}$
$\Rightarrow {\sin }^{-1}y=\frac{3\pi }{4}$
$\Rightarrow y\left(\frac{-1}{\sqrt{2}}\right)=\sin \left(\frac{3\pi }{4}\right)=\frac{1}{\sqrt{2}}$