Q. The differential equation for which $\sin^{- 1}{x + \sin^{- 1}{y = c}}$, is given by
A) $\sqrt{1 - x^{2}}dx + \sqrt{1 - y^{2}}dy = 0$
B) $\sqrt{1 - x^{2}}dy + \sqrt{1 - y^{2}}dx = 0$
C) $\sqrt{1 - x^{2}}dy - \sqrt{1 - y^{2}}dx = 0$
D) $\sqrt{1 - x^{2}}dx - \sqrt{1 - y^{2}}dy = 0$
Explanation
(b)
Given equation is
$\sin^{- 1}{x + \sin^{- 1}{y = c}}$ ….(i)
On differentiating Eq. (i) $w.r.t.x,$ we get
$\frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{1 - y^{2}}}\frac{dy}{dx} = 0$
$\Rightarrow \ \frac{dy}{dx} = - \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}}$
$\Rightarrow \ \sqrt{1 - x^{2}}\ dy + \sqrt{1 - y^{2}}dx = 0$
This is the required differential equation.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.