solution of differential equation $\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$ 
The solution of differential equation $\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$

$\bf{My\; Try::}$ Let $\displaystyle \frac{dy}{dx} = t\;,$ Then Diferential equation convert into $t^2-xt+y=0$
So Its solution is given by $\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}$
So we get $$\frac{dy}{dx} = \frac{x\pm \sqrt{x^2-4y}}{2}$$
Now How can I solve after that, Help me
Thanks

Here's another approach. Differentiating gives $y''\left(2y'-x\right)=0$, so $y''=0$ or $y'=\frac{x}{2}$. The former option gives $y=ax+b$ so $a^2-ax+ax+b=0$ and $b=-a^2$. The latter option gives $y=\tfrac{x^2}{4}+c$ so $\frac{x^2}{4}-\frac{x^2}{2}+\tfrac{x^2}{2}+c=0$, which work iff $c=0$. The solution is $y=ax-a^2$ or $y=\frac{x^2}{4}$.
