A) \[2x{{e}^{{{\tan }^{-1}}y}}={{e}^{2\,{{\tan }^{-1}}y}}+c\]
B) \[x{{e}^{{{\tan }^{-1}}y}}={{\tan }^{-1}}y+c\]
C) \[x{{e}^{2\,{{\tan }^{-1}}y}}={{e}^{{{\tan }^{-1}}y}}+c\]
D) \[(x-2)=c{{e}^{-{{\tan }^{-1}}}}y\]
Correct Answer: A
Solution :
Given differential equation is \[(1+{{y}^{2}})+(x-{{e}^{{{\tan }^{-1}}}}y)\frac{dy}{dx}=0\] \[\Rightarrow \] \[(1+{{y}^{2}})\frac{dx}{dy}=-x+{{e}^{{{\tan }^{-1}}y}}\] \[\Rightarrow \] \[\frac{dx}{dy}+\frac{x}{1+{{y}^{2}}}=\frac{{{e}^{{{\tan }^{-1}}}}y}{1+{{y}^{2}}}\] Which is a linear differential equation. Here, \[P=\frac{1}{1+{{y}^{2}}},Q=\frac{{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}\] \[IF={{e}^{\int{P\,\,ly}}}={{e}^{\int{\frac{1}{1+{{y}^{2}}}dy}}}={{e}^{{{\tan }^{-1}}y}}\] \[\therefore \] Solution is \[x.IF\,=\int{Q.IF\,\,dy+c}\] \[x\,{{e}^{{{\tan }^{-1}}y}}=\int{\frac{{{e}^{{{\tan }^{-1}}}}}{1+{{y}^{2}}}}.{{e}^{{{\tan }^{-1}}y}}+\frac{c}{2}\] \[\Rightarrow \] \[x\,{{e}^{{{\tan }^{-1}}y}}=\frac{{{e}^{2\,{{\tan }^{-1}}y}}}{2}+\frac{c}{2}\] \[\therefore \]\[2x\,\,{{e}^{{{\tan }^{-1}}y}}={{e}^{2\,{{\tan }^{-1}}y}}+c\]
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