Question

The differential equation corresponding to the family of circles given by $(x-a{)}^2+(y-b{)}^2=4$ where a and b are parameters, is

• A. $4\frac{d^{2}y}{d{x}^2}+9y=0$
• B. $4{\left(\frac{d^{2}y}{d{x}^2}\right)}^2={\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3$
• C. $4\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=6y$
• D. $4{\left(\frac{d^{2}y}{a{x}^2}\right)}^2+{\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^2=0$

Answer 1

The correct option is B $4{\left(\frac{d^{2}y}{d{x}^2}\right)}^2={\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3$

${\left(x-a\right)}^2+{\left(y-b\right)}^2=4$ $\left(1\right)$

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

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

$1+\left(y-b\right)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0\left(3\right)$

$-\left(y-b\right)\frac{dy}{dx}=x-a$ using $\left(2\right)$
putting it in $\left(2\right)$

${\left(y-b\right)}^2\left[{\left(\frac{dy}{dx}\right)}^2+1\right]=4$ $\left(4\right)$

$\left(y-b\right)\frac{d^{2}y}{d{x}^2}=-\left[{\left(\frac{dy}{dx}\right)}^2+1\right]$ using $\left(4\right)$

${\left(y-b\right)}^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2={\left[{\left(\frac{dy}{dx}\right)}^2+1\right]}^2$
putting this in $\left(4\right)$

${\left[{\left(\frac{dy}{dx}\right)}^2+1\right]}^3=4{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$