If the solution curve y = y ( x ) of the differential equation (1 +...

Given: $\left(1 + {y}^{2}\right)\left(1 + \log_{e}x\right)dx + xdy = 0$

$⇒ \dfrac{\left(1 + \log_{e}x\right)}{x}dx = \dfrac{- 1}{\left(1 + {y}^{2}\right)}dy$

$⇒ ∫ \left(\dfrac{1}{x} + \dfrac{\log x}{x}\right)dx + ∫ \dfrac{dy}{1 + {y}^{2}} = 0$

$⇒ \log x + \dfrac{( \log x{)}^{2}}{2} + \tan^{- 1}y = C$

This curve passes through the point \left(1, 1\right)$.</p><p>$⇒ \log \left(1\right) + \dfrac{( \log \left(1\right){)}^{2}}{2} + \tan^{- 1}\left(1\right) = C$</p><p>$⇒ C = \dfrac{\pi}{4}$</p><p>$⇒ \log x + \dfrac{( \log x{)}^{2}}{2} + \tan^{- 1}y = \dfrac{\pi}{4}$</p><p>Now, putting$x = e$</p><p>$⇒ loge + \dfrac{( loge{)}^{2}}{2} + \tan^{- 1}y = \dfrac{\pi}{4}$</p><p>$⇒ \tan^{- 1}y = \dfrac{\pi}{4} - \dfrac{3}{2}$</p><p>$⇒ y = \tan \left(\dfrac{\pi}{4} - \dfrac{3}{2}\right)$</p><p>$⇒ y = \dfrac{tan\dfrac{\pi}{4} - \tan \left(\dfrac{3}{2}\right)}{1 + tan\dfrac{\pi}{4}\tan \left(\dfrac{3}{2}\right)}$</p><p>$⇒ y = \dfrac{1 - \tan \left(\dfrac{3}{2}\right)}{1 + \tan \left(\dfrac{3}{2}\right)}$</p><p>Hence, on comparing we get,</p><p>$⇒ \alpha = \beta = 1$</p><p>$⇒ \alpha + 2\beta = 3$