A) \[y=-\frac{1}{2}{{\tan }^{-1}}x+c\]
B) \[y+\log x+\frac{{{x}^{2}}}{2}+c=0\]
C) \[y=\frac{1}{2}{{\tan }^{-1}}x+c\]
D) \[y-\log x-\frac{{{x}^{2}}}{2}=c\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}+\frac{1+{{x}^{2}}}{x}=0\] Þ \[dy+\left( \frac{1}{x}+x \right)\text{ }dx=0\] On integrating, we get \[y+\log x+\frac{{{x}^{2}}}{2}+c=0\].You need to login to perform this action.
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