If the solution curve of the differential equation $((\tan^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point $(1, 0)$ then the abscissa of the point on the curve whose ordinate is $\tan (1)$ is :

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2 $e$

$\frac{2}{e}$

$\frac{1}{e}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{\tan^{- 1} y}{1 + y^{2}} \\ I.f = e^{\int \frac{1}{1 + y^{2}} d y} = e^{\tan^{- 1} y} \\ s o l u t i o n i s \\ x e^{\tan^{- 1} y} = \int \frac{\tan^{- 1} y}{1 + y^{2}} e^{\tan^{- 1} y} d y \\ x \cdot e^{\tan^{- 1} y} = (\tan^{- 1} y - 1) e^{\tan^{- 1} y} + c \\ ∵ s u b s t i t u t i n g (1, 0) ⇒ c = 2 . \\ For y = \tan 1 \Rightarrow x = \frac{2}{e} \end{array}$