A) \[y{{\tan }^{-1}}x=c\]
B) \[x{{\tan }^{-1}}y=c\]
C) \[y+{{\tan }^{-1}}x=c\]
D) \[x+{{\tan }^{-1}}y=c\]
Correct Answer: A
Solution :
\[ydx+(1+{{x}^{2}}){{\tan }^{-1}}xdy=0\] Þ \[\int_{{}}^{{}}{\frac{dx}{(1+{{x}^{2}}){{\tan }^{-1}}x}}=-\int_{{}}^{{}}{\frac{dy}{y}}\] Þ \[\frac{1}{{{2}^{x}}}-\frac{1}{{{2}^{y}}}=\frac{{{c}_{1}}}{\log 2}=c\] Þ \[\log (y{{\tan }^{-1}}x)+\log c=0\]Þ\[y{{\tan }^{-1}}x=c\].
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