Question

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

The correct option is C

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

Explanation for the correct option:

Step-1 : Solution of differential equations:

Given, $\sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0,\left|x\right|<1$

On solving, we get:

$$\begin{gathered} \sqrt{1-x^2}\frac{dy}{dx}+\sqrt{1-y^2}=0 \\ \Rightarrow \frac{dy}{\sqrt{1-y^2}}+\frac{dx}{\sqrt{1-x^2}}=0 \end{gathered}$$

On integrating,

$$\begin{gathered} \Rightarrow \int \frac{dy}{\sqrt{1-y^2}}+\int \frac{dx}{\sqrt{1-x^2}}=0 \\ \Rightarrow {\sin }^{-1}x+{\sin }^{-1}y=c.....\left(1\right) \\ Atx=\frac{1}{2},y=\frac{\sqrt{3}}{2} \\ \Rightarrow {\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)+{\sin }^{-1}\left(\frac{1}{2}\right)=c \\ \Rightarrow \frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}=c \\ \Rightarrow \frac{\mathrm{\pi }}{2}=c \\ \Rightarrow c=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Step-2: Value of $y\left(\frac{-1}{\sqrt{2}}\right)$:

Substituting the value of $c$ in equation (1), we get:

$$\begin{gathered} {\sin }^{-1}y+{\sin }^{-1}x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\sin }^{-1}y=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x={\cos }^{-1}x\mathbf{[}\mathbf{\because }{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{]} \end{gathered}$$

Now,

$\begin{array}{rcl}y\left(-\frac{1}{\sqrt{2}}\right)& =& \sin \left({\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)\\ & =& \frac{1}{\sqrt{2}}\end{array}$

$\therefore y\left(-\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$

Hence, the correct answer is option (C).