A curve $y=f\left(x\right)$ has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of the point P from the x-axis. Then the differential equation of the curve 

Equation of normal

$$\begin{gathered} Y-y=-\frac{1}{m}(X-x)\Rightarrow -my+mY=X-x \\ \begin{array}{l}x+mY=X+my\\ X-mY+my-x=0\end{array} \end{gathered}$$

Perpendicular from $(0,0)=\left|\frac{x+my}{\sqrt{1+m^2}}\right|=y\Rightarrow x^2+2xym=y^2$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}\Rightarrow$  homogeneous

Also,  $x.2y.\frac{dy}{dx}-x^2=y^2$

Put,  $y^2=t; 2y\frac{dy}{dx}=\frac{dt}{dx};x.\frac{dt}{dx}+x^2=t$

$\frac{dt}{dx}-\frac{1}{x}t=-x$  Which is linear differential equation.
Hence, (a). (b) and (d) are the correct answers