Question

Find the differential equation of the family of circles ${\left( {x - a} \right)^2} - {\left( {y - b} \right)^2} = {r^2}$ .

Answer 1

Hint : The given differential equation for the family of circles should be differentiated two times, say concerning to x as it consists of two arbitrary constants in the form of a and b.

Complete step-by-step answer :
The given differential equation of family of circles is
${\left( {x - a} \right)^2} - {\left( {y - b} \right)^2} = {r^2} \cdots \left( 1 \right)$
The equation consists of two arbitrary constants. In order to obtain the differential equation for such type should be differentiated two times.
Differentiating equation (1) concerning to x , we get
$2\left( {x - a} \right) - 2\left( {y - b} \right)\dfrac{{dy}}{{dx}} = 0 \cdots \left( 2 \right)$
On simplifying the equation (2), we get
$\left( {x - a} \right) - \left( {y - b} \right)\dfrac{{dy}}{{dx}} = 0 \cdots \left( 3 \right)$
Differentiating equation (3) concerning to $x$ again we get,
$1 - \dfrac{{dy}}{{dx}} \times \dfrac{{dy}}{{dx}} - \left( {y - b} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 0 \\\ 1 - {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = \left( {y - b} \right)\dfrac{{{d^2}y}}{{d{x^2}}} \\\ \left( {y - b} \right) = \dfrac{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}}{{\dfrac{{{d^2}y}}{{d{x^2}}}}} \cdots \left( 4 \right) \\\$
Substitute the value of $\left( {y - b} \right)$ in equation (3), we get
$\left( {x - a} \right) - \dfrac{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}}{{\dfrac{{{d^2}y}}{{d{x^2}}}}} \times \dfrac{{dy}}{{dx}} = 0 \\\ \left( {x - a} \right) = \dfrac{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]\left[ {\dfrac{{dy}}{{dx}}} \right]}}{{\dfrac{{{d^2}y}}{{d{x^2}}}}} \cdots \left( 5 \right) \\\$
Substitute the value of and from equation (5) and (4) in equation (1), we get
$\Rightarrow \dfrac{{{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}^2}{{\left[ {\dfrac{{dy}}{{dx}}} \right]}^2}}}{{{{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]}^2}}} + \dfrac{{{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}^2}}}{{{{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]}^2}}} = {r^2} \cdots \left( 6 \right)$
Taking out $\dfrac{{{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}^2}}}{{{{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]}^2}}}$ common in equation (6) to simplify the equation, we get
$\Rightarrow \dfrac{{{{\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]}^2}}}{{{{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]}^2}}}\left[ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} + 1} \right] = {r^2} \\\ {\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]^2}\left[ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} + 1} \right] = {r^2}{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]^2} \\\$
Hence the required solution of the family of circles
$\Rightarrow {\left( {x - a} \right)^2} - {\left( {y - b} \right)^2} = {r^2}$ is ${\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]^2}\left[ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} + 1} \right] = {r^2}{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]^2}$ .

Note : The important thing in order to obtain the differential equation is to eliminate the arbitrary constants present in the original equation.
The equation should be differentiated as many times as there are constants present in the original equation. For instance if the number of constants is one it should be differentiated once,etc.
The order of the differential equation is the order of the highest derivative present in the equation and the degree of the differential equation is the degree of the highest order derivative.
In the differential equation ${\left[ {1 - {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]^2}\left[ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} + 1} \right] = {r^2}{\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]^2}$ calculated above , the order and degree both are equal to 2.