Question

The differential equation satisfing ${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}c,$ where $c$ is an arbitrary constant, is

• A. $\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$
• B. $\frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$
• C. $\frac{dy}{dx}=-\frac{\sqrt{1-x^2}}{\sqrt{1-y^2}}$
• D. $\frac{dy}{dx}=\frac{\sqrt{1-x^2}}{\sqrt{1-y^2}}$

Answer 1

The correct option is B $\frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Given, ${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}c$
Now differentiating both sides with respect to $x$ we get, $\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.
This is the required differential equation.