If a curve passes through the origin and the slope of the tangent...

$y\left(0\right) = 0$

and $\dfrac{dy}{dx} = \dfrac{{\left(x - 2\right)}^{2} + y + 4}{x - 2}$

$⇒ \dfrac{dy}{dx} - \dfrac{y}{x - 2} = \left(x - 2\right) + \dfrac{4}{x - 2}$

$⇒ I . F . = {e}^{- ∫ \dfrac{1}{x - 2}dx} = \dfrac{1}{x - 2}$

Solution of L.D.E.

$⇒ y\dfrac{1}{x - 2} = ∫ \dfrac{1}{x - 2}\left(\left(x - 2\right) + \dfrac{4}{x - 2}\right) \cdot dx$

$⇒ \dfrac{y}{x - 2} = x - \dfrac{4}{x - 2} + C$

Now, at $x = 0, y = 0 ⇒ C = - 2$

$y = x\left(x - 2\right) - 4 - 2\left(x - 2\right)$

$⇒ y = {x}^{2} - 4x$

This curve passes through \left(5, 5\right)$