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