# find the differential equation of the family of straight lines at a fixed distance p from the origin?

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.

edited Jun 15, 2013 by meena.p