The solution of $(1 + y^{2}) + (x - e^{\tan - 1} y) y_{1} = 0$ is

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$(x - 2) = {Ke}^{- {Tan}^{- 1} y}$

$2 {xe}^{{Tan}^{- 1} y = e^{2 Tan - 1} y} + k$

${xe}^{2 {Tan}^{- 1}} y = e^{{Tan}^{- 1}} + k$

${xe}^{Tan - 1} y = {Tan}^{- 1} + k$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n (1 + y^{2}) = (e^{\tan^{- 1} y} - x) \frac{d y}{d x} \\ \Rightarrow \frac{d x}{d y} = \frac{e^{\tan^{- 1} y}}{1 + y^{2}} - \frac{1}{1 + y^{2}} x \\ \Rightarrow \frac{d x}{d y} + \frac{1}{1 + y^{2}} x = \frac{e^{\tan^{- 1} y}}{1 + y^{2}} \\ I . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} \\ S o l u t i o n i s \\ x (I . F) = \int \frac{e^{\tan^{- 1} y}}{1 + y^{2}} (I . F) d y \\ \Rightarrow x . e^{\tan^{- 1} y} = \int \frac{e^{\tan^{- 1} y}}{1 + y^{2}} e^{\tan^{- 1} y} d y \end{array}$

$\begin{array}{l} L e t \tan^{- 1} y = v \\ \Rightarrow \frac{1}{1 + y^{2}} d y = d v \\ x e^{\tan^{- 1} y} = \int e^{2 v} d v \\ = \frac{1}{2} e^{2 v} + c \\ = \frac{1}{2} e^{\tan^{- 1} y} + c \\ \Rightarrow 2 x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + k \end{array}$