If a curve passes through the origin and the slope

$\frac{d y}{d x}=\frac{(x-2)^{2}+y+4}{(x-2)}=(x-2)+\frac{y+4}{(x-2)}$

Let $x-2=t \Rightarrow d x=d t$

and $y+4=u \Rightarrow d y=d u$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dt}}$

$\frac{d u}{d t}=t+\frac{u}{t} \Rightarrow \frac{d u}{d t}-\frac{u}{t}=t$

I. $\mathrm{F}=\mathrm{e}^{\int \frac{-1}{t} \mathrm{dt}}=\mathrm{e}^{-\mathrm{Int}}=\frac{1}{t}$

u. $\frac{1}{t}=\int t \cdot \frac{1}{t} d t \Rightarrow \frac{u}{t}=t+c$

$\frac{y+4}{x-2}=(x-2)+c$

Passing through $(0,0) c=0$

$\Rightarrow(y+4)=(x-2)^{2}$