**Toolbox:**

- General equation of a straight line passing through the origin is $y = mx$
- Equation of a straight line which is at a fixed distance of $p$ from the origin is $xcos \theta +ysin \theta = p$

$\mathbf{Step 1:}$

Let $ x \cos \theta + y \sin \theta = p$ -----(1)

On differentiating with respect to $x$ we get

$cos\theta + \large\frac{dy}{dx}$$ \sin \theta = 0$

This implies $\large\frac{dy}{dx} $$=-\cos \theta$

or $\cos \theta =\large\frac{- dy}{dx}$$[\sin \theta]$

Substituting for $\cos \theta$ in equ(1) we get

$ x. \large\frac{-dy}{dx}$$[ \sin \theta] + y \sin \theta = p $-----(2)

$\mathbf{Step 2:}$

Now differentiating equ (2) again with respect to x we get.

$x \large\frac{dy}{dx}$$[\sin \theta]$ can be differentiated by applying product rule.

$x.\large\frac{d^2y}{dx^2}$$[\sin \theta] + (-1) \large\frac{dy}{dx} $$[\sin \theta] + \large\frac{dy}{dx}$$[\sin \theta] = 0$

Cancelling $\sin \theta$ throughout we get,

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

$\qquad\quad\quad x.\large\frac{ d^2y}{dx^2}$$ = 0$

$x,y'' = 0$ is the required differential equation.