Question

Obtain the differential equation of all circles of radius r and centre $\left( {0,0} \right)$

Answer 1

Use the general equation of the circle that is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$. In this question the centre is (0,0) so h and k are zeros. Substitute the zeros and differentiate the obtained equation by using product rule.

Complete Step by Step Solution:
The objective of the problem is to obtain the differential equation of all circles of radius r and centre
$\left( {0,0} \right)$.
Let us assume the general equation of the circle ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
It is given that the centre is at $\left( {0,0} \right)$ and the radius is r. So in the general equation h,k will be zero and the radius is the same.
Now substitute zeroes in the place of h and k.
Therefore, ${\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {r^2} \\\ {x^2} + {y^2} = {r^2}.........\left( 1 \right) \\\$
Equation 1 represents the equation of the circle with centre $\left( {0,0} \right)$and radius r.
Now we have to obtain a differential equation of equation 1. For this differentiate equation 1 with respect to x on both sides.
Now take equation 1 and differentiate on both sides with respect to x.
$\dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \dfrac{d}{{dx}}\left( {{r^2}} \right)$
Using additive property of differentiation that is $\dfrac{d}{{dx}}\left( {u + v} \right) = \dfrac{d}{{dx}}\left( u \right) + \dfrac{d}{{dx}}\left( v \right)$ we get
$\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {{r^2}} \right)$
Now using the differentiation formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ we get
$2x + 2y\dfrac{{dy}}{{dx}} = 2r$
Taking two as common we get$x + y\dfrac{{dy}}{{dx}} = r$
Again differentiate above equation on both sides we get
$\dfrac{d}{{dx}}\left( {x + y\dfrac{{dy}}{{dx}} = r\dfrac{{dr}}{{dx}}} \right)$
By using product rule of differentiation on term second term of above equation we get
$\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {y\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( r \right) \\\ \Rightarrow 1 + y\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) + \dfrac{{dy}}{{dx}}\left( {\dfrac{d}{{dx}}\left( y \right)} \right) = 0 \\\ \Rightarrow 1 + y\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 0 \\\$

$1 + y\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 0$ is the required differential equation of circle.

Note:
The product rule of differentiation is$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. There are so many standard formulae to solve the differentiation simply and easily. A differential equation is an equation involving derivatives of a function. We should derive the general equation until the constants were removed from the general equation.