Differential Equations Question 36
Question: The general solution of the differential equation $ ydx+(1+x^{2}){{\tan }^{-1}}xdy=0, $ is
[MP PET 1995]
Options:
A) $ y{{\tan }^{-1}}x=c $
B) $ x{{\tan }^{-1}}y=c $
C) $ y+{{\tan }^{-1}}x=c $
D) $ x+{{\tan }^{-1}}y=c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ ydx+(1+x^{2}){{\tan }^{-1}}xdy=0 $
Therefore $ \int _{{}}^{{}}{\frac{dx}{(1+x^{2}){{\tan }^{-1}}x}}=-\int _{{}}^{{}}{\frac{dy}{y^{2}}} $
Therefore $ \frac{1}{2^{x}}-\frac{1}{2^{y}}=\frac{c}{\log 2} $
Therefore $ \log (y{{\tan }^{-1}}x)+\log c=0 $
Therefore $ y = c \tan^{-1}x $ .
BETA
