If the solution curve of the differential equation (dy)/(dx) = (x +...

Given,

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

Now let $x - 1 = X, y - 1 = Y$

So, $\dfrac{dy}{dx} = \dfrac{X + Y}{X - Y} . . . . . . . . \left(1\right)$

Now let $Y = VX\dfrac{dY}{dX} = V + X\dfrac{dV}{dX}$

Putting the value in equation \left(1\right)$we get,</p><p>$V + X\dfrac{dV}{dX} = \dfrac{1 + V}{1 - V}$</p><p>$⇒ X\dfrac{dV}{dX} = \dfrac{{V}^{2} + 1}{1 - V}$</p><p>$⇒ ∫ \dfrac{1 - V}{1 + {V}^{2}}dV = ∫ \dfrac{dX}{X}$</p><p>$⇒ ∫ \dfrac{dV}{1 + {V}^{2}} - \dfrac{1}{2} ∫ \dfrac{2VdV}{1 + {V}^{2}} = ∫ \dfrac{dX}{X}$</p><p>$⇒ \tan^{- 1}V - \dfrac{1}{2}\ln \left(1 + {V}^{2}\right) = \ln X + c$</p><p>$⇒ \tan^{- 1}\left(\dfrac{Y}{X}\right) - \dfrac{1}{2}\ln \left(1 + \dfrac{{Y}^{2}}{{X}^{2}}\right) = \ln \left(X\right) + c$</p><p>$⇒ \tan^{- 1}\left(\dfrac{y - 1}{x - 1}\right) - \dfrac{1}{2}\ln \left(1 + \dfrac{{\left(y - 1\right)}^{2}}{{\left(x - 1\right)}^{2}}\right) = \ln \left(x - 1\right) + c$</p><p>Now given curve passes through \left(2, 1\right)$

So, $0 - \dfrac{1}{2}\ln 1 = \ln 1 + c ⇒ c = 0$

Now given curve also passes through \left(k + 1, 2\right)$</p><p>So,$\tan^{- 1}\left(\dfrac{1}{k}\right) - \dfrac{1}{2}$In \left(1 + \dfrac{1}{{k}^{2}}\right) = lnk$

$⇒ 2\tan^{- 1}\left(\dfrac{1}{k}\right) =$ In \left(\dfrac{1 + {k}^{2}}{{k}^{2}}\right) + 2lnk$</p><p>$⇒ 2\tan^{- 1}\left(\dfrac{1}{k}\right) =$In \left(1 + {k}^{2}\right)$