Differential equation of a parabola with vertex at origin and symmetric about x-axis is :

$x\frac{dy}{dx}-2y=0$ $2x\frac{dy}{dx}-y=0$ $x\frac{d^2y}{dx^2}-2\frac{dy}{dx}=0$ $2\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$

$2x\frac{dy}{dx}-y=0$

The correct answer is Option (2) → $2x\frac{dy}{dx}-y=0$ $y^2=4ax$  ....(1) from (1) $\frac{y^2}{x}=4a$ differentiating wrt x $2y\frac{dy}{dx}=4a$ $2y\frac{dy}{dx}=\frac{y^2}{x}$ so $2x\frac{dy}{dx}-y=0$