If the solution curve of the differential equation (( ^ - 1 y) -...

\left(\tan^{- 1}y - x\right)dy = \left(1 + {y}^{2}\right)dx$</p><p>$⇒ \dfrac{dx}{dy} + \left(\dfrac{1}{1 + {y}^{2}}\right)x = \dfrac{\tan^{- 1}y}{1 + {y}^{2}}$</p><p>I.F.$= {e}^{∫ \dfrac{1}{1 + {y}^{2}}dy} = {e}^{\tan^{- 1}y}$</p><p>General solution will be</p><p>$x\left({e}^{\tan^{- 1}y}\right) = ∫ {e}^{\tan^{- 1}y} \times \dfrac{\tan^{- 1}y}{1 + {y}^{2}}dy$</p><p>$⇒ x\left({e}^{\tan^{- 1}y}\right) = \tan^{- 1}y\left({e}^{\tan^{- 1}y}\right) - {e}^{\tan^{- 1}y} + C$</p><p>Since the curve passes through \left(1, 0\right) ⇒ C = 2$

So, the curve reduces to $x\left({e}^{\tan^{- 1}y}\right) = \tan^{- 1}y\left({e}^{\tan^{- 1}y}\right) - {e}^{\tan^{- 1}y} + 2$

when $y = \tan 1, x\left(e\right) = 1\left(e\right) - e + 2$

$⇒ x = \dfrac{2}{e}$