Differential Equations Question 138
Question: The differential equation for which $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=c $ is given by
[Karnataka CET 2003]
Options:
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 $
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ {{\sin }^{-1}}x+{{\sin }^{-1}}y=c $
…..(i) On differentiating w.r.t. to x, we get $ \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}=0 $
Therefore $ \frac{dy}{dx}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} $
Therefore $ \sqrt{1-x^{2}}dy+\sqrt{1-y^{2}}dx=0 $ .
BETA
