Differential Equations Question 355
Question: The solution of the equation $ \frac{dy}{dx}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0 $ is
[Orissa JEE 2003]
Options:
A) $ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}}=c $
B) $ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=c $
C) $ x\sqrt{1+y^{2}}+y\sqrt{1+x^{2}}=c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{dy}{dx}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0 $
Therefore $ \int _{{}}^{{}}{\frac{dy}{\sqrt{1-y^{2}}}}=-\int _{{}}^{{}}{\frac{dx}{\sqrt{1-x^{2}}}} $
Therefore $ {{\sin }^{-1}}y=-{{\sin }^{-1}}x+{{\sin }^{-1}}c $
Therefore $ {{\sin }^{-1}}[ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} ]={{\sin }^{-1}}c $
Therefore $ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=c $ .
BETA
