Question

The order, degree of the differential equation of all circles of radius r, having centre only axis passing through the origin is (where r is arbitrary constant)
A.1,1
B.2,1
C.3,1
D.4,2

Answer 1

Hint : We need to find the order and degree of a differential equation of a circle. We need to remember the equation of the circle and then by differentiating the circle equation at origin. Then we can find the order and degree of the differential equation. Note that the differential equation is an equation which involves derivatives.

Complete step-by-step answer :
The order of a differential equation is the order of the highest order derivative present in the differential equation. The degree of a differential equation is the exponent of the highest order derivative present in the differential equation.
Now, the standard equation of the circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$ with centre $(h,k)$ .
Let $(0,a)$ be the centre of the circle and radius $a$ , then above becomes,
$\Rightarrow {(x - 0)^2} + {(y - a)^2} = {a^2}$
Using ${(a - b)^2} = {a^2} + {b^2} - 2ab$ and cancelling ${a^2}$ terms on both sides.
Then we have, $\Rightarrow {x^2} + {y^2} - 2ay = 0$ -- (1)
Differentiate with respect $y$ .
$\Rightarrow 2x \dfrac{{dx}}{{dy}} + 2y - 2a = 0$
$\Rightarrow 2x \dfrac{{dx}}{{dy}} = 2( - y + a)$
By the definition of order and degree of the differential equation, we can see that order is 1, degree is 1.
So, the correct answer is “Option A”.

Note : We can also differentiate above with respect to x, it will confuse while taking the common of differentiative terms. So, we have two y terms and one x term in (1) we can differentiate with respect to y there is no need to take common and rearranging. Either way you get the same answer. Remember the definition of order and degree you can solve for any problem.