A) \[(x-2)=k{{e}^{-{{\tan }^{-1}}y}}\]
B) \[2\,x\,{{e}^{-{{\tan }^{-1}}y}}={{e}^{2\,{{\tan }^{-1}}y}}+k\]
C) \[x{{e}^{{{\tan }^{-1}}}}={{\tan }^{-1}}\,y+k\]
D) \[x{{e}^{2\,\,{{\tan }^{-1}}y}}={{e}^{{{\tan }^{-1}}y}}+k\]
Correct Answer: B
Solution :
\[(1+{{y}^{2}})+(x-{{e}^{{{\tan }^{-1}}y}})\frac{dy}{dx}=0\] \[\Rightarrow \] \[(1+{{y}^{2}})\frac{dx}{dy}+x={{e}^{{{\tan }^{-1}}y}}\] \[\Rightarrow \] \[\frac{dx}{dy}+\frac{1}{1+{{y}^{2}}}x=\frac{{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}\] \[IF={{e}^{\int_{{}}^{{}}{P\,dy}}}={{e}^{\int_{{}}^{{}}{\frac{1}{1+{{y}^{2}}}dy}}}={{e}^{{{\tan }^{-1}}y}}\] Therefore, required solution is \[x\,.\,\,{{e}^{{{\tan }^{-1}}y}}=\int{{{e}^{{{\tan }^{-1}}y}}.\frac{{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}dy+{{k}_{1}}}\] \[\Rightarrow \] \[x{{e}^{{{\tan }^{-1}}y}}=\int_{{}}^{{}}{\frac{{{e}^{2{{\tan }^{-1}}y}}}{1+{{y}^{2}}}\,dy+{{k}_{1}}}\] \[\Rightarrow \] \[x{{e}^{{{\tan }^{-1}}y}}=\frac{1}{2}{{e}^{2{{\tan }^{-1}}y}}+{{k}_{1}}\] \[\Rightarrow \] \[2\,x\,{{e}^{{{\tan }^{-1}}y}}={{e}^{2{{\tan }^{-1}}y}}+k\]
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