If a curve passes through the origin and the slope of the tangent to it at any point x ,   y is x 2 - 4…

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

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

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

$\Rightarrow I\mathit{.}F.=e^{-\int \frac{1}{x-2}dx}=\frac{1}{x-2}$

Solution of L.D.E.

$\Rightarrow y\frac{1}{x-2}=\int \frac{1}{x-2}\left(\left(x-2\right)+\frac{4}{x-2}\right)·dx$

$\Rightarrow \frac{y}{x-2}=x-\frac{4}{x-2}+C$

Now, at $x=0,y=0\Rightarrow C=-2$

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

$\Rightarrow y=x^2-4x$

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