Differential Equations Question 340
Question: If $ \frac{dy}{dx}+\frac{1}{\sqrt{1-x^{2}}}=0 $ , then
[MNR 1983]
Options:
A) $ y+{{\sin }^{-1}}x=c $
B) $ y^{2}+2{{\sin }^{-1}}x+c=0 $
C) $ x+{{\sin }^{-1}}y=0 $
D) $ x^{2}+2{{\sin }^{-1}}y=1 $
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Answer:
Correct Answer: A
Solution:
$ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}} $
Therefore $ dy=-\frac{1}{\sqrt{1-x^{2}}}dx $
On integrating, we get $ y={{\cos }^{-1}}x+c $
Therefore $ y=\frac{\pi }{2}-{{\sin }^{-1}}x+c $
Therefore $ y+{{\sin }^{-1}}x=c $ .
BETA
