A solution of the differential equation, ${\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}+y=0$ is :

Given${\left(\frac{dy}{dn}\right)}^2-x\frac{dy}{dn}+y=0$
$\Rightarrow y=x\left(\frac{dy}{d_n}\right)-{\left(\frac{dy}{dn}\right)}^2$
Differentiating both sides of above equation w.r.t  $x$,

$\Rightarrow \frac{dy}{dn}=\frac{dy}{dx}+x{\left(\frac{dy}{d{w}^2}\right)}^{a_1}-2\left(\frac{dy}{dn}\right)\left(\frac{d_y^2}{d{n}^2}\right)$ 

$\Rightarrow 0=\left(x-2\left(\frac{dy}{dx}\right)\right)\left(\frac{d^{2}y}{d{n}^2}\right)$
$\frac{d^{2}y}{d{n}^2}=0 or x=2\frac{dy}{dn}$
$\Rightarrow \frac{dy}{dn}=c$ (constant) This doesn't satisfy given
$\Rightarrow y=cx+d$ ($i$) equation.
But, given : $y=x\frac{dy}{dn}-{\left(\frac{dy}{dn}\right)}^2$ …….$\left(1\right)$
$\therefore$ Comparing $\left(i\right)$ and $\left(ii\right)$, we get,
$\frac{c}{\left(\frac{dy}{dn}\right)}=\frac{d}{{\left(\frac{dy}{dn}\right)}^2}$
$\Rightarrow \frac{dy}{dn}=-\frac{d}{c}=c$
$$\begin{gathered} \Rightarrow d=-c^2 \\ \Rightarrow y=cn-c^2 ......\left(\mathrm{iii}\right) \end{gathered}$$

From given option only $y=2n-4$ matches

general solution $\left(iii\right)$ where $C=2$