If a curve passes through the origin and the slope of

Given

$y(0)=0$

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

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

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

Solution of L.D.E.

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

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

Now, at $x=0, y=0 \Rightarrow C=-2$

$y=x(x-2)-4-2(x-2)$

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

This curve passes through $(5,5)$