Question

The order of the differential equation of all circles having radius $r$ is.

• A. One
• B. two
• C. three
• D. four

Answer 1

The correct option is B two

The equation of family of circle is $(x-a{)}^2+(y-b{)}^2=r^2$ ---- ( 1 )

Clearly, the given equation have two arbitrary constants, so we differentiate twice w.r.t $x$

$\Rightarrow$ $2(x-a)+2(y-b)=0$

$\Rightarrow$ $(x-a)+(y-b)\frac{dy}{dx}=0$

$\Rightarrow$ $(x-a)=-(y-b)\frac{dy}{dx}$ ---- ( 2 )

Differentiate again w.r.t $x,$

$1+(y-b)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$

$\Rightarrow$ $(y-b)=-\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}$ ---- ( 3 )

Substitute ( 3 ) and ( 2 ) in ( 1 ), we get,

$\Rightarrow$ ${\left[\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}\frac{dy}{dx}\right]}^2$$+\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^2}{\frac{d^{2}y}{d{x}^2}}=r^2$

$\Rightarrow$ ${[1+{\left(\frac{dy}{dx}\right)}^2]}^2[1+{\left(\frac{dy}{dx}\right)}^2]=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$

$\Rightarrow$ ${[1+{\left(\frac{dy}{dx}\right)}^2]}^3=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$

$\therefore$ We can see required order is $2$

as the Order of a differential equation is the order of the highest derivative present in the equation.