Question

The order and degree of the differential equation of the family of circles of fixed radius $r$ with centres on the $ y-axis $ , are respectively.
(a) $2,2$
(b) $2,3$
(c) $1,1$
(d) $3,1$
(e) $1,2$

Answer 1

Hint: For solving this question first we will find the equation of the family of circles of fixed radius $r$ with centres on the $y-axis$ and then we will differentiate it to get its differential equation. After that, we will see the basic definition of the order and degree of the differential equation to give the correct answer easily.

Complete step-by-step solution -
Given:
We have to find the order and degree of the differential equation of the family of circles of fixed radius $r$ with centres on the $y-axis$ .
Now, as we know that the general equation of the circle is $C:{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ whose centre is at point $\left( a,b \right)$ and radius of the circle is $r$ . And as it is given that centre of the circle is on the $y-axis$ . So, we can take the centre of the circle at point $\left( 0,b \right)$ and radius $r$ . Then,
$C:{{x}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$
Now, ${{x}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ is the equation of the family of circles of fixed radius $r$ with centres on the $y-axis$ so, we will differentiate its equation with respect to $x$ . Then,
$\begin{align}
  & {{x}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}} \\
 & \Rightarrow \dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( {{\left( y-b \right)}^{2}} \right)}{dx}=\dfrac{d\left( {{r}^{2}} \right)}{dx} \\
\end{align}$
Now, we will write $\dfrac{d\left( {{x}^{2}} \right)}{dx}=2x$ , $\dfrac{d\left( {{\left( y-b \right)}^{2}} \right)}{dx}=2\left( y-b \right)\dfrac{dy}{dx}$ and $\dfrac{d\left( {{r}^{2}} \right)}{dx}=0$ in the above equation. Then,
$\begin{align}
  & \dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( {{\left( y-b \right)}^{2}} \right)}{dx}=\dfrac{d\left( {{r}^{2}} \right)}{dx} \\
 & \Rightarrow 2x+2\left( y-b \right)\dfrac{dy}{dx}=0 \\
 & \Rightarrow \left( y-b \right)\dfrac{dy}{dx}=-x \\
 & \Rightarrow \left( y-b \right)=-\dfrac{x}{\left( \dfrac{dy}{dx} \right)} \\
\end{align}$
Now, put $\left( y-b \right)=-\dfrac{x}{\left( \dfrac{dy}{dx} \right)}$ in the equation $C:{{x}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{\left( -\dfrac{x}{\left( \dfrac{dy}{dx} \right)} \right)}^{2}}={{r}^{2}} \\
 & \Rightarrow \dfrac{{{x}^{2}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}={{r}^{2}}-{{x}^{2}} \\
 & \Rightarrow \dfrac{{{x}^{2}}}{{{r}^{2}}-{{x}^{2}}}={{\left( \dfrac{dy}{dx} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{dy}{dx} \right)}^{2}}-\dfrac{{{x}^{2}}}{{{r}^{2}}-{{x}^{2}}}=0 \\
\end{align}$
Now, the above equation is the differential equation of the family of circles of fixed radius $r$ with centres on the $y-axis$ . But we should see the definitions of “Order of a Differential Equation” and “Degree of a Differential Equation” to answer correctly.
Order of a Differential Equation:
The order of a differential equation is the order of the highest order derivative appearing in the equation. For example, in the following differential equation:
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+3\dfrac{dy}{dx}+2y=0$
The order of the highest derivative is 2. So, it is a differential equation of order 2. And the equation $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-6{{\left( \dfrac{dy}{dx} \right)}^{4}}-4y=0$ is of order 3 because the order of the highest order derivative in it is 3.
Moreover, we should note that the order of a differential equation is a positive integer.
Degree of a Differential Equation:
The degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients. For example, in the following differential equation:
$\dfrac{{{d}^{3}}y}{d{{x}^{3}}}-6{{\left( \dfrac{dy}{dx} \right)}^{4}}-4y=0$
In this equation power of the highest order derivative is 1. So, it is a differential equation of degree one.
 Now, we come back to our problem in which we have got ${{\left( \dfrac{dy}{dx} \right)}^{2}}-\dfrac{{{x}^{2}}}{{{r}^{2}}-{{x}^{2}}}=0$ as the differential equation of the family of circles of fixed radius $r$ with centres on the $y-axis$ . And from the above discussion, it is evident that its order will be 1 and degree will be 2.
Hence, (e) will be the correct option.

Note: Here, the student should first understand what is asked in the question and after that proceed in the right direction to get the correct answer quickly. Moreover, we should form the equation of the family of the circles carefully as per the given data and don’t commit mistakes while differentiating the equation of the family of circles and write the differential equation correctly. After that, don’t confuse between degree and order of the differential equation while giving the final answer.