A) \[{{\tan }^{-1}}y\]
B) \[{{e}^{{{\tan }^{-1}}y}}\]
C) \[\frac{1}{1+{{y}^{2}}}\]
D) \[\frac{1}{x(1+{{y}^{2}})}\]
Correct Answer: B
Solution :
\[(1+{{y}^{2}})dx-({{\tan }^{-1}}y-x)dy=0\] Þ \[\frac{dy}{dx}=\frac{1+{{y}^{2}}}{{{\tan }^{-1}}y-x}\] Þ \[\frac{dx}{dy}=\frac{{{\tan }^{-1}}y}{1+{{y}^{2}}}-\frac{x}{1+{{y}^{2}}}\] Þ \[\frac{dx}{dy}+\frac{x}{1+{{y}^{2}}}=\frac{{{\tan }^{-1}}y}{1+{{y}^{2}}}\] This is equation of the form \[\frac{dx}{dy}+Px=Q\] So, I.F. \[={{e}^{\int{P\,dy}}}={{e}^{\int{\frac{1}{1+{{y}^{2}}}.dy}}}={{e}^{{{\tan }^{-1}}y}}\].
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