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

Given: $\dfrac{dy}{dx} = \dfrac{x + y - 2}{x - y}$

Now, finding the intersection point of $x + y - 2 = 0$ and $x - y = 0$.

$⇒ x + x - 2 = 0, x = y$

$⇒ x = 1, y = 1$

$⇒ \left(x, y\right) ≡ \left(1, 1\right)$

Let, $x = X + 1, y = Y + 1$

$⇒ \dfrac{dY}{dX} = \dfrac{X + Y}{X - Y}$

Putting, $Y = tX$

$⇒ \dfrac{dY}{dX} = t + X\dfrac{dt}{dX}$

$⇒ \dfrac{1 + t}{1 - t} = t + X\dfrac{dt}{dX}$

$⇒ \dfrac{1 + t - t + {t}^{2}}{1 - t} = X\dfrac{dt}{dX}$

$⇒ \dfrac{1}{X}dX = \dfrac{1 - t}{1 + {t}^{2}}dt$

$⇒ ∫ \dfrac{1}{X}dX = ∫ \dfrac{1 - t}{1 + {t}^{2}}dt$

$⇒ \log \left|X\right| = \tan^{- 1}t - \dfrac{1}{2}\log \left|1 + {t}^{2}\right| + C$

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

It is given that the solution curve passes through \left(2, 1\right)$.</p><p>$⇒ \log \left|2 - 1\right| = \tan^{- 1}\left(\dfrac{1 - 1}{2 - 1}\right) - \dfrac{1}{2}\log \left|1 + {\left(\dfrac{1 - 1}{2 - 1}\right)}^{2}\right| + C$</p><p>$⇒ 0 = \tan^{- 1}\left(0\right) - \dfrac{1}{2}\log \left|1\right| + C$</p><p>$⇒ C = 0$</p><p>$⇒ \alpha = 1, \beta = 2$</p><p>$⇒ 5\beta + \alpha = 11$.