Differential Equations Question 21
Question: The solution of the differential equation $ xy\frac{dy}{dx}=\frac{(1+y^{2})(1+x+x^{2})}{(1+x^{2})} $ is
[AISSE 1983]
Options:
A) $ \frac{1}{2}\log (1+y^{2})=\log x-{{\tan }^{-1}}x+c $
B) $ \frac{1}{2}\log (1+y^{2})=\log x+{{\tan }^{-1}}x+c $
C) $ \log (1+y^{2})=\log x-{{\tan }^{-1}}x+c $
D) $ \log (1+y^{2})=\log x+{{\tan }^{-1}}x+c $
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Answer:
Correct Answer: B
Solution:
$ xy\frac{dy}{dx}=\frac{(1+y^{2})(1+x+x^{2})}{(1+x^{2})} $
Therefore $ \int _{{}}^{{}}{\frac{ydy}{1+y^{2}}=\int _{{}}^{{}}{\frac{(1+x+x^{2})}{x(1+x^{2})}}}dx=\int _{{}}^{{}}{\frac{1}{x}dx}+\int _{{}}^{{}}{\frac{dx}{1+x^{2}}} $
Therefore $ \frac{1}{2}\log (1+y^{2})=\log x+{{\tan }^{-1}}x+c $ .
BETA
