Question

The differential equation whose solution is ${(x-h)}^2+{(y-k)}^2=a^2$ is $(a$ is a constant $)$:

• A. ${[1+{\left(\frac{dy}{dx}\right)}^2]}^3=a^2\frac{d^{2}y}{{dx}^2}$
• B. ${[1+{\left(\frac{dy}{dx}\right)}^2]}^3=a^2{\left(\frac{d^{2}y}{{dx}^2}\right)}^2$
• C. ${[1+\left(\frac{dy}{dx}\right)]}^3=a^2{\left(\frac{d^{2}y}{{dx}^2}\right)}^2$
• D. None of these

Answer 1

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

Given differential eqn is
${(x-h)}^2+{(y-k)}^2=a^2$ .....(1)
Differentiating (1) w.r.t. x ,
$2(x-h)+2(y-k)\frac{dy}{dx}=0$
$(x-h)+(y-k)\frac{dy}{dx}=0$ ....(2)
$\Rightarrow \frac{dy}{dx}=-\frac{x-h}{y-k}$ ....(3)
Squaring both sides of eqn (3), we get
${\left(\frac{dy}{dx}\right)}^2=\frac{(x-h{)}^2}{(y-k{)}^2}$
Adding $1$ to both sides, we get
$\Rightarrow 1+{\left(\frac{dy}{dx}\right)}^2=\frac{a^2}{(y-k{)}^2}$
$\Rightarrow (y-k{)}^2=\frac{a^2}{1+{\left(\frac{dy}{dx}\right)}^2}$ ....(4)
Differentiating (2) w.r.t. x, we get
$1+(y-k)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$
Substituting the value of $y+k$ from (4), we get
$1+\frac{a}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$
$\Rightarrow 1+{\left(\frac{dy}{dx}\right)}^2=-\frac{a}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}\frac{d^{2}y}{d{x}^2}$
$\Rightarrow {(1+{\left(\frac{dy}{dx}\right)}^2)}^{3/2}=-a\frac{d^{2}y}{d{x}^2}$
Squaring both sides, we get
${[1+{\left(\frac{dy}{dx}\right)}^2]}^3=a^2{\left(\frac{d^{2}y}{{dx}^2}\right)}^2$